# Applied Optimization Problems (Procedures and Examples)

Video Series: Calculus 1 (Explore, Discover, Learn) Series

(D4M) — Here is the video transcript for this video.

00:00
[Music] it’s time for some optimization problems optimizing with numbers
volume geometry area angles distance time and even marginal analysis
in this video i present you an optimization procedure
and then take you through it with each of these examples
all right welcome to the episode of applied optimization problems
procedures and examples in this video i’m going to start us off with the

00:01
optimization procedure that we’re going to uh use and then i’m going to
go through lots of examples of working out this optimization procedure and then
after we get good at that we’ll talk about some a marginal
analysis we’ll work on some business calculus type of problems
um and then at the end we’ll work through some
we’ll talk about some exercises so let’s get started
so you know when when you’re starting uh to work on an application problem
or sometimes they’re called word problems um of course you want to
understand the problem and you want to be asking questions like
what are the unknowns like what what is it that i don’t know
and what is it that i do know what is given to to us in the problem

00:02
and then the next step is as usually to draw a picture
try to get some kind of intuition about what’s going on
and you want to label the unknowns and then you want to start to try to
introduce notation you know label where’s the x’s where are the y’s
and what are those represent are those representing miles or speed or you know
whatever you want to label introducing notation
of course the final step after you work out through the whole procedure
is you want to look back at your original
uh problem re-read it and make sure that it makes sense
for the answer that you’re supplying and so you know just that kind of uh broad
drawing pictures try to make it more abstract by introducing notation
and then you get to use the calculus and then you want to go back and check
your work all right so here’s the step-by-step call

00:03
you know we want to use derivatives and so the first step is often
sketch or graph or you know draw some kind of diagram
introduce mathematical notation this step is important because
you don’t want to just start using l what is l what is v is v volume or is v
velocity right so you want to make sure that you
introduce notation and by that i mean don’t just start writing notation
actually say what it is what does x represent what does y
represent and then you want to come up with equations
expressions and functions that you can work with
and then you want to formulate the problem mathematically
by stating what you know and what you don’t know
and then you want to uh identify theorems and validate hypotheses so some common
theorems that we’ve worked on so far is the uh first derivative test

00:04
concavity test um extreme value theorem is a huge one
and so whatever theorem you’re going to use you need to make sure first that you
can use it and it won’t give you wrong results so i think step
e is a little bit different than say pre-calculus
we can pre-calculus was it so important that you
stated what theorem you’re using and that you needed to validate some
hypothesis i mean you know that’s not obviously a strict rule but i think that
in calculus you’re going to see that emphasized a great deal more
you know obviously depends on where you’re taking your course and
what class you have and what your instructor is actually requiring you to do
but i think that this is a big step that a lot of calculus teachers will say
you know what theorem did you use okay sure you took a derivative
everyone knows that you’re going to take a derivative but what theorem did you
use with that derivative all right and so then obviously you gotta

00:05
find that derivative and apply that theorem and you gotta solve the problem
right you maybe do some additional algebra whatever you need to do
and then you gotta state the problem back in terms of the original language
and the important thing here i think is especially important is units
if the problem gives you miles per hour then your answer should be
also in miles per hour and you know you should use the original
you’re multilingual and someone comes up to you and asks you
a question in a language then you often reply back in that language right
so in any case there’s a good optimization preach
procedure there and let’s get started on some examples [Music]
okay so i thought we would get started optimizing volume first

00:06
so this is a nice immediate problem that is going to illustrate a lot for us
and we will be able to um get a lot of knowledge from this problem here in
terms of the overall procedure and so let’s look at how to do this problem here
find the dimensions of a right circular cylinder of largest volume
that can be inscribed in a sphere of radius capital r
so notice first here that you that this um capital r is given to us they give us
the sphere and we’re asking um what’s the largest
circular cylinder that can be inscribed in there
and so even though we have a you know a picture on the
on the right there you know it um it’s good to try to graph or draw one mice

00:07
ourselves so let’s try to try to do that here so let’s say we have a sphere
so let’s say this is a sphere and let’s draw this around here
and we need the center of the sphere so i’ll put that in there and then we need
to inscribe a right circular cylinder so this is this is a right circular
cylinder it’s not uh like bent over crooked it’s gone it’s up it’s up straight
it’s not like slanted right so i’m gonna make it right here
open it put it right here and it’s going to come up here [Music]
and then go through there and so there’s the center of the sphere
and now i need to start labeling things and now you know the thing that’s given
to us is capital r the radius of the sphere now the radius
of the spheres from here out to the end of the sphere now you
could put that radius anywhere you want but actually it touches here i’m going

00:08
to put it i’m going to put my radius here and so that length right there is is r
that’s r and so i need to be asking here like what is the
height of the um cylinder um now i’ll put an h here and i’m going
to put a little r here so little r is the radius of the
right circular cylinder so r is the radius [Music] of the cylinder
and uh 2h is the height right so here’s h and here’s another h
so 2h is the height of the cylinder [Music]
and you know this is a right circular cylinder this is a right angle right there

00:09
so we want to find r and you know we want to find 2h which is the height
and so we know the um volume of the sphere right so what is the volume of the
of the cylinder um it’s it’s going to be given by so this
i’ll just say here v is the volume [Music] of the cylinder [Music]
v is the volume of the capital v is the volume of the cylinder
and you know we know this relationship here r
squared plus h squared equals r squared from just the pythagorean theorem it’s a
right circular cylinder so we know that right there so now we
can try to write the volume as a function and so you know what is

00:10
the volume of the sphere equal to well it’s the area of the
top part right here which is pi r squared times the height which is 2h
now we would like this to be a function so we can take the derivative of it
so is there a way to get rid of the r squared here
and yes we can solve for r squared here so we’ll think of the volume as a
function of h and we’ll write it as 2 pi h so 2 pi h and then we have an r
squared and the r squared is just capital r squared minus the little r squared
okay so this is a function of h remember capital r is given to us that
is the radius of the sphere so this will give us a function right

00:11
here this is the volume of the cylinder in terms of h only and so
you know we have here the boundary and the boundary is 0 to r
and so i’m going to use the extreme value theorem here
so let’s put that over here extreme value theorem
um why because we have a continuous function so actually r squared is
r squared minus h squared that should be h squared here right
so as i was saying we have a function of h here
and this is continuous function of h and so we can use extreme value theorem
here and so you know h is bounded between zero and the
the sphere h cannot be any longer than the radius of the sphere so

00:12
um we are going to find the critical numbers so let’s find the derivative
[Music] so the derivative is what so we have
product rule here right so derivative of the first [Music]
plus two pi h times the derivative of the second derivative that part is zero
so times minus two h and so this derivative simplifies we have a
2 pi that factors out [Music] and let’s see what we have left we have
an r squared minus an h squared and then we have a minus two h squared
and so the derivative here is two pi r squared minus three h squares
all right very good and so where are we equal to zero

00:13
when this part right here is zero so let’s solve that over here r squared
minus three h squared is zero remember we’re solving for h
because we already know what capital r is they told us that that is a given to
us so capital r could be for example three thousand five thousand we don’t know
what capital r is but it’s given to us so solving this for h
so we have three h squared equals r squared or said differently h is what r over
square root of three and so we need to check uh v of zero b of r and v of
um each of the boundary boundary right here zero and
r so zero we need to check v of zero the volume when h is zero which is just
zero and what happens to when this is r is here
that’s also zero and so this will be r of square root of three and so we’ll plug

00:14
in here r over square 3 into our volume function right here
and when we do that we’re going to get out um the maximum
so the maximum is going to be given with by this um h value right here
and so you know we we’ve checked that this is continuous
and we found the critical number right here and so we’ve checked that this is
this is zero here and this is zero here and so
you know this right here must be a maximum by the extreme value theorem
and so now we can go and solve for r and so you know what would be the r
knowing this is the h so we can come up back up over here and solve for the r

00:15
and so the r would be the um r squared is r squared minus h
squared and so r squared would be r squared minus and then we’re going to
square this h so that would be r squared over three
um you know and so this would be what two thirds of r squared
two thirds r squared [Music] um don’t need the yeah and so r would be the what
um let’s just say we have 2 square roots of three over three and then
r you know to go ahead and rationalize that
when we solve this for r we’re going to take square root all right so the um

00:16
let’s go ahead and put the here the sorry and then so this is the height right
here is the r over square root of three and
we need the uh actually that’s not the height the height is 2h
so that this will be the height here we’ll just put put this as the height
is a 2r over square root of 3 and then if we solve for r here this will be
what square root of 6 if we rationalize over 3 r and so
this will be the height and this will be the radius we’ll call this the radius
so those are the dimensions the radius and the height so just in terms of
the original problem we want to state what the height of it is

00:17
so it’s two h’s so and then and then the radius here um and now we can you know
see if this makes any sense or not um of course it should satisfy this
equation right here and um because they don’t give us the r
that we don’t actually have decimal approximations for that
so it’s kind of hard to know if that makes any sense um but you know this
is um what we get so um this will be the dimensions
and so there’s an example of using the extreme value theorem here
it’s important to notice that we have a closed bounded interval here
and that we’re continuous on it so we can find our critical numbers here
all right very good so let’s go on to the next one
and now we’re going to optimize uh perimeter

00:18
so let’s read this problem here very carefully um a woman plans to fence off a
rectangular garden and we have the given area so the area is 64 feet squared
which should be the dimensions of the garden if she wants to
minimize the amount of fencing to be used
um so i’m just going to say like this is x and this is y
and so we have 64 is x times y is the area and we want to know what are
the dimensions of the guard if she wants to minimize the amount of fencing used
alright so we’re actually looking for the amount of fencing used so that’s
perimeter so the perimeter will be 2x plus 2y
um but you know we want this to be a function
of either x or y so let’s make this a function of

00:19
x so i’m going to say this is 2x plus 2 and then for the y i’m going to use 64
over x [Music] so there’s our perimeter which we want to try to minimize
we’re trying to minimize the amount of fencing to be used so i want to minimize
this right here and so you know the domain here is
x is greater than zero this is our perimeter function
so let’s find our derivative here so p prime is so we have two
plus and then what was that uh 128 and then we have a minus sign so minus 128
over x squared and then here getting common denominator

00:20
it’s 2x squared minus 128 over x squared and we notice that the derivative here
is continuous on its domain right here and so you know we’re just looking for
when the derivative is zero that’ll be our only critical number
um so you know this is going to be you know something we could solve for uh
pretty easily 2x squared minus 128. you can kind of see that
x is just eight two times 64 right um so x squared is um 128 over two
so x is plus or minus eight but only eight is going to be in our domain here
so our only shot here is is at eight now because we’re not having a closed
bounded interval here necessarily um so one approach would be to try to use the
first or second derivative test so i’m going to apply the first derivative test
so i’m going to look at the interval 0 to 8 and i’m going to see what’s

00:21
happening at 8 and then we’ll see what’s happening greater than eight
so i’m gonna look at the function and the derivative and i’m gonna make some
conclusion here so when i test a number say between zero
and eight i’m going to be testing it say up here
um for example i have let’s say a one and then the minus here is going to
dominate and that’s going to be positive so this will be negative and so that
tells me it’s decreasing [Music] and when i try something greater than 8
for example 100 now this one is dominating so that would be 100
squared and then minus 128 so that’ll be positive over positive
that’s positive now it’s increasing so for decreasing and then we’re increasing
this is going to be a relative min [Music] there’s a relative min happening at
eight now this is the first derivative test

00:22
now if we look at this problem though what’s happening is that
this function is is decreasing between zero and eight
and then increasing forever after eight is always
increasing so it looks something like this it’s going down and it’s going up
you know on zero to eight it’s decreasing and then it’s increasing so this right
here is actually an absolute max so because of the domain of the function
and this information right here so x equals eight um so we have an absolute max
absolute max at x equals eight [Music] from from looking at this all the
information right here and then um so at at x equals eight we have an absolute
max and i’m sorry absolute men i keep saying absolute max

00:23
absolute min that’s absolute min um and then so the dimensions of the
of the um right because it’s asking um what should the dimensions of the
garden be right so when x is eight we just come back up here and say yeah y
is y is eight also so the dimensions are x equals eight and then what are we
going to say here feet and y equals eight feet and that shouldn’t be
much of a surprise it’s just going to be a square [Music]
just a square an eight foot by eight foot square [Music]
so let’s look at the next example this time we’re going to be optimizing
an angle [Music] and we are going to read this very carefully so says the

00:24
bottom of an eight foot muriel painted on a vertical wall is 13
foot above the ground so let’s say we have 13 feet here
and this is eight feet above the ground so let’s say there’s another eight feet
here and then um the lens of a camera fixed to a tripod
is four foot above the ground so let’s say this is four foot here
and how far from the wall should the camera be placed
so i’m going to say this right here is x i could put the x here but i’m going to
put the x here and how far from the wall should the camera be placed
to photograph the mirror with the largest possible angle
so here’s the angle that we’re interested in

00:25
this angle right here that’s supposed to be a straight thing so there’s a little
bit more straight so i’m going to make some angles here to to get to this angle
inside here um i’m looking at these two triangles here
so i’m going to look at this angle here of the larger triangle
and call that angle alpha and i’m going to look at the angle of the smaller
triangle and call that beta and i’m going to look at this angle here
and call that angle theta so theta is the alpha larger angle
minus the smaller angle take away the beta and so we’re trying to maximize this
alpha here i’m sorry we’re trying to maximize the theta here
the angle in which we need to photograph the uh painting

00:26
the mural so we need to to maximize theta um so i’m going to be using
the inverse tangent in order to get to these
angles here so when i’m looking at the small triangle right here
you know this this triangle right here maybe we can
the small triangle right here where where we have a beta
right here and then so what is this length right here
what is that length right there right so we know this is x and what is this
length right here so this whole thing is 13 and then we have a 4 out here
so this is 9 right here that’s a 9 right there so this theta can be written as
this alpha minus beta and and what is the alpha right here
so the alpha is 17 right so this would be tangent inverse of 17 over x now
okay wait where did i get that from right so if you just look at the

00:27
triangle right here what do we know about tangent of alpha tangent of alpha is
the side over here 17 over the x the opposite over the
adjacent that’s just 17 over x right so to get to the alpha i use tangent
inverse similarly what is tangent of beta tangent of beta is the smaller one
nine over the x so tangent of beta is nine over x so to get to that beta i
just take arctan of both sides right so this will be minus beta so minus
arctan and for the beta we’re going to use 9 over the x [Music]
so this is the angle in terms of the x this is alpha of x here
so alpha is the angle of elevation from the camera lens to the top

00:28
and beta is the angle of elevation from the camera to the bottom of the mural
so we have this function right here and we’re going to go and try to take
the derivative of this angle right here we don’t know um other than x is greater
than zero and so you know let’s take the derivative here and see what we get
so the derivative of theta with respect to x here is
so remember the derivative of arctan is one over
one plus whatever this is squared times the derivative of this right here so
the derivative of that is what um minus 17 over x squared [Music]
and then we have a minus sign right here and now the derivative of the arctan
uh tangent inverse of nine over x so that’ll be one

00:29
over one plus whatever this is nine over x squared [Music]
times the derivative of nine over x which is minus nine over x squared
okay and let’s try to simplify this a little bit
now this x squared here is going to be distributed through
and so what we’re looking at here is minus 17 over we have an x squared
plus 17 squared minus and then we have a minus 9 right here
so this will be plus and this will be 9 over
now this x squared can be distributed through here and it’ll be x squared
and then we’ll distribute it here we’ll get the 9 squared
and so i like to clear those fractions with that x squared that’s
that’s a sweet thing to do right there that simplifies it

00:30
a lot but we still need to get a common denominator so we’re going to multiply
minus 17 and then we’re going to get an x square plus
an 81 and then plus 9 times an x squared plus right so [Music]
17 squared is what 289 [Music] and then this is all over x squared plus 289
times the x squared plus 81. okay so we have the x squared plus 17 squared
and the x squared plus 9 squared and now we have a common denominator here
so let’s see what we’re going to get for this numerator
we’re going to get a minus 17x squared and then a positive
x squared here so we’re going to get a minus eight x squared and we’re gonna get

00:31
minus seventeen times eighty one plus a nine times two eighty nine and
for that we’re gonna get a positive 153 if you do that calculation
and so we get x square plus 289 times x squared plus 81.
okay and so then we need to go and solve this equal to zero to find the
critical numbers and we will get actually that will be
let’s see here minus 17 times 81 times 81 plus nine times 289

00:32
calculate all that up and we’re going to get a um it’s not 153 it’s 153 times
eight so that’ll be 42 so that’d be 12. so i think it’s 12 24. [Music]
all right so we’ll be able to factor an 8 out of this on top here in fact i’m
going to just factor out a minus eight and that’ll be x square and this will be
minus and there’s the 153 and then there’s an x square plus 289
and then times x squared plus 81. all right so the whole point is that in
order to finish solving this we need to set x squared minus 153
equal to zero so x is square root of 153 [Music]
or if you want to break that down it’s three square roots of 17.

00:33
and that is approximately so 3 square root of 17 is approximately
12.4 and this will be in feet [Music] so we want to go x needs to be about
12.4 feet and then that angle theta will be maximized
so how far from the wall should the camera be placed
12.4 feet and that will give us the largest possible angle there
now in order to finish this problem we need to take this x value right here
three square roots of 17 and we need to apply the second derivative test and
here’s the second derivative right here and what we’ll check is that
this is giving us the largest possible angle so if we set up a table
less than 3 square roots of 17 and check greater than 3 times the square

00:34
root of 17 and we check the derivative on these intervals here
then what we’ll get is a positive and a negative
and we can check the derivative right here so something greater than for example
say a hundred and then if we do like 100 squared
this will be positive so that negative will make it all negative
and then you know check this right here and you get positive
so it’s increasing and then it’s decreasing and at 3 00 to 17
we have a relative max but if you look at the domain which is just
uh greater than zero strictly greater zero it’s going to be
increasing until it gets to that x value and then it’s going to be decreasing
forever more after that this is greater than so this relative

00:35
max actually has to be an absolute max and so using the first derivative test
and that argument right there that we only have in increasing and then
decreasing we can argue that this is a relative maximum right here
uh this we can argue that this is an absolute maximum
so you know if you have increasing decreasing increasing decreasing and so on
your relative maximum may not always be an absolute maximum
but in this case because it’s increasing and then it’s always decreasing after
that you can argue that this relative max is in fact an absolute max
so for this x right here that will give us the largest possible angle now
i’m sorry yeah that would give us the largest possible angle but this is the
x this is how far it is away from the wall
all right so there’s optimizing an angle um usually optimizing an angle not
always but i would um you know try to use some trig functions

00:36
whether it’s tangent or sine or cosine whatever
but because i was given the opposites oh and i was uh
and i wanted to involve the x which is which is the adjacent
so i decided to use the tangents okay so next example [Music]
optimizing areas next so let’s read this problem here
so someone with a 700 foot fence wants to enclose a rectangular area and
divide it into four pins so let’s look how to do that [Music]
so we’re dividing it into four pins so now we have one
two three four so we have four pins and what is the largest possible total area
of the four pins so we’re trying to do the area here and we know that we have

00:37
a 750 feet of fencing that’s a fixed number let’s call this x and this y and
let’s say we have here the perimeter or which is the the fencing
is 700 and what is that going to be one two three four five so that’ll be five x
plus two y so we have that expression right there
in terms of the amount of fencing um and then the area so what will the area be
so the area the four pins is just x times y and so what are we going to try to
maximize right so what is the largest possible area so
we need to try to maximize area so i’m going to say this is a of x which is
x times the y um actually what is the y so if i move 5x over and then divide by

00:38
2 so i’m going to get 750 minus 5x all divided by 2. that’s that’s what the
y is so this is just x times y and now we want to try to find the largest total
area so we want to try to maximize the area there so let’s just say here that
the you know x is greater than zero so let’s go here with the derivative of a is
um maybe we can simplify this a little bit first this is what um 750 over 2
right so that is going to be what it’s still got another fraction there so

00:39
we’ll put 750 over 2 x minus 5 over 2 x squared right and so what is 750
over two that is just simply um so seven 375 right [Music] let’s put that say
over here 375 um minus five over two [Music] okay so [Music]

00:40
the derivative is um just simply [Music] the the so let’s
um when we solve this for y yeah okay so the derivative is just
simply going to be sorry this is missing an x here that was throwing me off here
and then this is minus 5 over 2 x squared okay sorry so this is 375 minus 5x
and we want to look where that’s zero so we’re just going to stop for that x
so the x is so the x is going to be you know move the five
x over and then divide by five so 750 divided by five

00:41
and let’s see what is that that is going to be that’s going to be [Music] 75
and so when i’m looking at this x is 75 here and then i can go find the y
so we found that x here and then [Music] this will be the critical number right
where the derivative is zero so now we can go find the largest possible area
it’ll be a times um the x 75 times what the 750 minus five times the 775
and then all divided by two um so 75 times that number right there

00:42
and then we’re going to get 14 0 625 and this will be square feet [Music]
so we can go and you know just plug that into your calculator you get
14 and 62.5 square feet and so that’s of course the area and the x was the 75
here and that’s where the derivative is zero and now you know
you want to claim that this is the largest possible total area
so we would look at this derivative right here
at 75 right here we can look at it less than 75
we can look at it greater than 75 and we can check the

00:43
derivative and we can make our conclusion here we can do the
first derivative test here and so when you test this right here out
you’ll get the less than 75 for example you know something grade 0
but for example 1 right so that’ll be positive
and greater than 75 for example 100 that’ll be negative right so it’s
increasing and it’s decreasing so that’s at 75 we’re going to get a max
it’s going to be a relative max but because after greater 35 it’s always
decreasing it’s never going to get greater it’s never going to start increasing
so this right here this relative max is is actually going to be an absolute max
at 75 so that x is 75 there will give us the
um greatest largest possible area so we’re using the first derivative test

00:44
but we’re actually using more than that we’re actually looking at the conclusion
of the first derivative test and we’re making an argument that that has to be
the maximum the absolute maximum all right so next example [Music]
all right let’s look at some geometry so find all points on the circle
such that the product of the x and y coordinate is this largest possible [Music]
so here we’re looking at a circle of say radius a so i’ll just draw a
circle there and the radius is a now when we look at a point here on the
circle x and y if we’re looking in the first quadrant then x times y
will be positive so we want this to be as large as possible
so we don’t want any negative numbers right x

00:45
times y will be positive but also in the third quadrant
x times y will be positive because x and y are both negative and so that
product will be positive so we’re looking such as the product
x and y is as large as possible so we’re trying to maximize
this function right here let’s call it f of x and that’s going to be
x times y we want to maximize this maximize x times y
now we have an expression for the y we have x squared plus y squared equals a
squared now i’m going to restrict the attention of just looking at the
quadrant one and so y will be um square root of a squared minus x squared and so
so that’s just looking in the in the first quadrant there and so the function

00:46
that we’re trying to maximize now will look like f of x will be x times y
and i’m looking here you know in the first quadrant x and y are in quadrant one
in quadrant one so we’ll we’ll look at quadrant three at the end
um but for quadrant one this is what we have we’re trying to maximize this
function right here so let’s look at the derivative [Music]
so we have the one times the second and then plus now x and then we have a
one-half and then we have a [Music] fraction right to the minus one half
power and then times a minus two x so that’s the derivative there a squared
minus x squared and then all of that right so think of this as

00:47
to the one half power so one half comes down and then we have minus one half
power and then times the derivative of the inside part here all right so good
so this will probably simplify a little bit this will be uh square root
and then the twos are going to cancel and so we’re getting minus
x squared on top here so we’re going to get a minus 2 over 2
so we get a minus x squared on top and now if we
get a common denominator so we want a squared minus x squared
and then another a squared minus x squared all over
a squared minus x squared in other words think of that as over one
and multiply top and bottom by what we need to get a common denominator

00:48
so this i’ll just leave the same [Music] so long story short
this is just a squared minus x squared and then we have another minus x squared
we have a squared minus two x squares all over the square root of a squared
minus x squared and so there’s our derivative right there
try to show all the work for everybody okay so there’s our derivative i’m going
to erase all this now and just take this derivative and put it up here
so this will be a squared minus 2x squared and what are the critical numbers
right what are the critical numbers for this derivative so we have um
you know a squared minus two x squared is equal to zero

00:49
and so you know solving that for x we have what uh divide by two so x squared is
um a squared over two or x is you know just a over square root of two
so there’s our critical number right there
and so we can go and substitute that in here
and find the y and when we do that let’s see here
what happens when you substitute that in a of
a over square root of two so this will be the
x which will be the a over square root of 2 times the y which is square root
of a squared minus the x squared and this will be if we simplify all this
this will be a squared over two [Music] so that’s the maximum

00:50
right there so it says find all points such that the product is as large as
possible so that’s the largest possible and what are the x there’s the x
and so we go find the y and the y is the same
a over square root of two and this is in quadrant one
and so we have the points here a over square root of two and a over
square root of two and in quadrant three we have negative a
over square root of two and negative a over square root of two so
because this will be you know x times y this will be the x and this
will be the y and when we multiply them the negative
times the negative we’ll get the same as if we multiplied these two right here
so that’s all points on the circle so we found one point over here and we

00:51
found one point over here and the that’s the x and y
coordinates for the two points that’s an a
okay so there’s some little little bit of geometry there [Music]
and let’s go on to the next one so let’s look at some numbers now
optimizing with numbers and so what we’re going to do is we’re
going to say find two non-negative numbers
and whose sum is eight and the product whose squares is as large as possible
so we want to maximize so let’s say we have two nine negative numbers so let’s
say x plus y is eight and they’re not negative
so x is greater than zero and y is greater than zero and we want the
product of the squares as large as possible

00:52
so the function which we want to be of only x is as large as possible
now you could make it in terms of y but i’ll just say choose x here um so
we want to you know do x squared times y squared we want to try to maximize this
and so i’ll say this is x squared times if we solve this for y it’s
eight minus x so x square times y squared and so now we can just say
x is greater than zero [Music] now let’s go ahead and take the derivative here
um of course you know one way to do it is just expand that out but
let’s go ahead and do this so we get 2x and then
eight minus x squared and then plus and then x squared and then a two eight
minus x and then a minus one [Music] and so let’s see if we can you know

00:53
get this looking good so what are we going to factor out each of them
so this has a two they both have two and so i can factor out a 2x
i can factor out a 2x and an 8 minus x so i’m going to take out that 2x and i
still have i still have an 8 minus x left here
so eight minus x now we took out that x we took out that two we still got an x
left still got a minus sign left and we took out that so i’m gonna say minus x
that’s eight eight minus x so this will be a two x sorry two x
and then eight minus x and this right here will be eight minus two x um

00:54
and so let’s go and take out 2 out and put that with there so let’s say this is
4x 8 minus x and then 4 minus x and so there’s our derivative looking
really nice there so the critical numbers are so critical numbers [Music]
um 0 4 and 8. now we want this to be as large as
possible we know nothing’s gonna happen from zero
right when x is zero then the product of x square plus
x squared times y square that product will be zero so
we’re really not really considering that to be a
possible answer but at this point we don’t know and so what is the largest
possible going to be um so we can look at um four and eight and see

00:55
which one gives us the um but we can make the first derivative
test we can say between four um between zero and four and equal to four
and between four and eight and equal to eight
and greater than eight and i can look at my derivative
and i can make a conclusion here [Music] and so what’s happening because we got
our derivative sitting right here so it’s really nice now first of all all
the x’s that we’re choosing here are positive so that’s always positive
right there now if i choose a number between zero and four like a one
and if i choose a one here and a one here then it’s all positive
if i choose a number between four and eight for example six
this is a six that’d be positive positive but that’ll be a negative there
and now if i choose a number greater than 8 for example a 10
then this will be negative and this will be negative so the whole thing will be

00:56
positive so increasing decreasing increasing so if it’s increasing and then
decreasing this is going to be a relative max [Music] and um you know
if it’s decreasing and increasing this is going to be a relative min
here to eight um and what’s happening at eight eight we’re going to get what uh
64 times so what will the y be the y will be zero
and that’s not going to give us anything useful either when the y is here when x
is a when the y is zero we multiply that product will be zero
so this is not going to give us anything here here either
so now we have the information here that we’re increasing
and then after four we’re going to be decreasing
um and until we get to eight but then you know x cannot be greater than eight so
you know one way to look at this is to maybe try to use extreme value theorem

00:57
on zero eight or you can try to use the first derivative test
and argue that this relative max has to actually be
an absolute max so this right here will be absolute max
at four and so now we can go find the the y the y will be of course four two
so absolute max [Music] at four four and what will be the actual
so find two non-negative numbers so we did that we found the x we found the y
so i’ll just say x is four and y is four and so the sum is eight and the product
of the squares as large as possible so you know sixteen times sixteen [Music]
that’s going to be the largest you can get if the sum is eight

00:58
okay so next example so another numbers problem
so here we go let’s see here now we have tried to minimize x squared y
so we’re going to try to minimize x squared y
and this is going to be our function let’s just call it
say p of x is going to be x squared y and then we can try to you know remove
that y so we have x squared and then what is the y so if i move the y over
and then you know divide by the 5 we’re going to get 2x minus 18 over 5.
okay so that’ll be our function right there and we’re
you know we know x is greater than or equal to zero we know y is greater than
or equal to zero and this is the function it’s trying to

00:59
minimize x squared y where we have a condition that we can solve for the y
now actually i want to take the time to talk about a little bit about calculus
three so once you make it past calculus one and calculus two
then you go into calculus 3. in calculus 3
you often will try to optimize problems but you won’t have a single variable you
often have multivariables so in calculus 3 you could learn how to
do this problem a different way using something called lagrange multipliers
but you know this is calculus one and so let’s do our one variable problem here
and so let’s take a derivative so we have here um [Music]
you know we could multiply this out or we could just do product rule [Music]

01:00
so let’s see here this will be two-fifths x to the third
and then um a minus eighteen-fifths x squared
and so we’ll have six fifths x minus 36 fifths x oh that’s the squared 36 and 36
fifths x [Music] and then we can factor out the one-fifth
um actually the six-fifths so six-fifths what do we get left the x-squared
minus the 6x so we should factor out the x also [Music]
six fix x and what we get left an x minus and then a six [Music]

01:01
so x equals zero from right here and x equals six
so setting the derivative equal to zero leads us to the critical numbers
so those are the two critical numbers now now we’re trying to minimize
x squared times y when x is zero that’s going to definitely give us um x squared
times y will be zero will we be able to get any smaller than that
well hold on when x is six maybe the y will be negative maybe we’ll get a
negative number out of that in fact what will be p of six
p of 6 will be 6 squared times the y but the y is the you know 2x
minus 18 over 5 and so if we calculate all that up we get minus 216 over five

01:02
and so that will be smaller than if we use the zero so when x is six
when x is six and y is the minus six fifths that we’re getting here
so then we have the minimum the minimum minimum of p so these two right here
give us the minimum of p the x squared times the y [Music] all right so
how do we know that at six we’re getting the
actual you know minimum value how we know there’s not one that’s smaller
so we would need to look at the behavior of the function
and we can do something like um less than zero
nope we’re not less than zero how about between zero and six

01:03
and then greater than six those are the interesting places
and we’re trying to look at this function right here
we’re trying to look at the derivative right here and so
we look at p prime we got the conclusion here
and so what’s happening between zero and six what’s happening for example a one
so positive negative so it’s decreasing and then greater than six for example
we’re like a ten that’s all positive so it’s decreasing and then it’s increasing
so at six we’re going to have so we’re going to have a relative min at six
and then using the fact that it’s always increasing when you’re
greater than six it’s always increasing in other words the graph is never going
to come back down and try to give you another minimum
so at x equals six we have a minimum so this is uh
using the first derivative test and then making the argument that that relative

01:04
min is in fact a global minimum so then at six there’s the value that we have
that is the actual minimum value and those are the x and y that give it
all right very good so let’s go on to the um interesting one here optimizing
distance [Music] this one’s a little bit long let’s see
if we can let’s see if we can solve this problem here
so let’s see here we have a truck it’s going 250 miles
due east of a sports car and it’s traveling
west at a constant speed so let’s see if we can encapsulate that
in a in a diagram so i’m going to draw with the car at the origin and the truck
at um away from here so here’s the car and here’s the truck right here and this

01:05
is at the point here 250 0 right because it’s 250 due east of a sports car
so here’s the car right here and this right here this truck is traveling west
and we know the speed that is traveling is traveling 60 miles per hour
meanwhile the sports car is going north at 80 miles per hour this would be 60
miles per hour and that’s 80 miles per hour
all right so there’s what’s happening right there now as this is going
north here um i’m going to say that the truck is
here i’m going to call this here x and i’m going to call this right here a y
[Music] so at time t the truck moved over here to x
and the sports car moved over here to y and so here’s the distance between them

01:06
and you know when will the truck and the car be closest to each other
so let’s call that distance a capital d make sure don’t use a lowercase d
and so we try to minimize the distance between them
so while the car is at the point here zero y and so let d be the distance that
separates them now this is the position right here this point right here is what
is 250 minus the x 0 right because it went that way
and so that’s 250 minus the x so the x is right here

01:07
so 250 and then minus the x right there [Music] so um x is the
distance right here [Music] yeah x is the distance the truck is gone
so this distance right here is 250 um minus the x right there and that will
give us this this part right in here and so
the distance let’s just put that over here to make make sure we’re good with
that so the distance is you know the distance squared pythagorean theorem right
and so the and so this will be the 250 minus x squared plus the y squared
so that would be the distance between them
so this part right in here will be the 250 minus the x
that’s that distance right there so the leg squared plus the leg squared equals

01:08
the hypotenuse squared there now we have here that dx dt is
60 and we have the dydt is 80 and so x is what 60 times t and the y
is 80 times t so 60 times t and the y is
the 90 80 times t so if you give me t is one second or sorry t is measured in
hours so if you give me one hour we’ve gone 60 miles okay so
when i’m looking at this d right here i’m going to try to
put it in terms of the t because we have two variables here
and we don’t have a way to you know make it in terms of only x’s
and y’s well i’m going to just simplify it and go to t

01:09
so this will be d squared is 250 minus 60 t
and then all that squared and then we’ll have the y
squared so 80t and then a squared and so now we want to simplify this and
just get a nice easy expression to work with for t so this
will be two hundred fifty sorry two thousand five hundred and then twenty five
minus twelve t plus four t squared and so you know expanding this out
simplifying it we get this and i want to take the derivative of this d
squared with respect to t so this will be easy to take the derivative with
respect to t so it’s worth um simplifying all this
and getting this right here because the derivative is now pretty easy to work

01:10
with what we’re gonna get is ten thousand times two t minus three
so the drift um the the derivative of the distance squared
is zero whenever we get this t right here
right so three over two or let’s say 1.5 and then so t was measured in hours so
let’s just say 1.5 hours here and this will produce the shortest distance
and so when will the truck and the in the car be the closest to each other
right so that’ll be when 1.5 hours and what is the minimum distance between them
so how do we find that minimum distance so we know that
when this t is 1.5 we can find the x and the y
so the minimum distance will be well let’s just go here with d squared is

01:11
the 250 minus the so 1.5 times the 80 so we’ll get 90 and then we’ll do a
squared plus and then we’re going to do the y squared so 1.5 times the 60.
so the 1.5 times the 60 we’re going to get the 120 squared [Music]
so this will be 1600 times the 25 and so when we take the square root
we’ll get 40 times 5 which is the 200 okay so this is the minimum distance

01:12
[Music] so that’s obviously measured in miles [Music] so
we use the extreme value theorem here on the interval 0 to
some distance right here the maximum distance
so we found the time and then we found the minimum distance what is the minimum
distance between them [Music] 220 miles all right very good
next one all right time for some [Music] marginal analysis
okay so let’s start off by talking about um some economic um concepts

01:13
um this should be you know every calculus student should
see a little physics should see a little business calculus just to have a good
idea of what’s going on in the world so marginal analysis is concerned with
way quantity such as price cost and revenue and profit
vary with small changes in production level
um right so you want to make tweaks you want to tweak things you always want to
tweak so you improve your revenue you want to
minimize your cost you want to find the perfect price so that you can maximize
your revenue and profit and so if we have a demand function that’s what
you really want to get to use your secret sauce there
your demand function that’s the price that consumers will pay for each unit
when x units are brought to the market so that’s
often found by doing a lot of market research

01:14
and if you know the demand function then you can find the total revenue
how much the price is the little p of x and times the number of units sold and
so that’ll be the amount of revenue and the profit capital p
we’re going to use as the revenue minus the cost
so this will be the total cost um you know the cost of producing it the cost
of marketing it you know the total cost involved
you know is being subtracted from the total revenue the amount of all the
revenue that you brought in for and that will give us the profit so c is
the total cost and so you know i’m going to use these
symbols throughout capital c capital r capital p and
we’re going to look at some examples where we talk about price cost

01:15
and profit so what is marginal analysis so you might remember
when we in an episode on the um linearization of a function
we talked about the linear approximation formula so we had something like
f of x plus delta x minus f of x is approximately equal to the derivative at
x times delta x and this is whenever delta x is close to zero
i’ll just say is small and this of course assuming that the
function is differentiable um and you know we use this
um what i call the linear approximation formula
we use this in a previous episode to do things like

01:16
um approximate some decimals and we used uh linearization which is based upon
this to find a linear function which represents another function
um on some interval and so you know look back at that episode to refresh the
we’re going to look at this the approximation formula
in terms of profit revenue and cost so r is for revenue um now
here we’re saying when delta x is small and so i think i i kind of remember that
episode i was arguing you know you know being small is
very relative term here because you know for example you might say is too

01:17
small or is too close to zero well i’m or if you made a 2 on an exam well the 2
is not pretty small in any case when you’re working with you know
business calculus or marginal analysis the x’s represent the number of
units um and so you know what’s a change in number of units right
so you’re not going to be making 0.3 of a hat
or half of a car right so you’re talking about x is whole numbers here so
the the the smallest delta x is gonna be is one
long story short so what would the revenue be
when this is one you know what is the linear approximation form is saying
when the delta x is one it represents one unit
so this will be approximately equal to r of x times delta x delta x is one

01:18
so r of x is called the marginal revenue and we can approximate the marginal
revenue right here using this right here and what this right here is
this right here represents the revenue from one additional unit
additional so this is the revenue of producing
x plus one units this is the revenue of producing
x units so if you subtract you’re just getting the revenue from producing
one extra unit and you can approximate that
using the derivative and that’s pretty actually amazing
if you want to know the price of the revenue of producing just one more
unit you can look at the revenue derivative of the revenue you can look
at the marginal revenue now you can do the same thing for cost

01:19
the cost of producing one additional unit this is the cost of
producing one additional unit is approximately equal to the marginal cost
so the derivative at x this is called the marginal cost
so the cost of producing one additional unit is approximately equal to the
marginal cost and similarly profit the profit from producing
one additional unit can be approximated by the marginal profit marginal profit
[Music] that’s marginal and so this is like the
most basic thing there is in marginal analysis
um at least from a calculus point of view so we have marginal revenue marginal
cost and marginal profit and those are the derivatives and what they represent

01:20
is an approximation for one additional unit all right so let’s work out some
problems now so let’s look at this problem here optimizing revenue and
what we’re going to do now is figure out this revenue function here
so a company mines low grade nickel ore and the company mines x
tons of ore if you can sell the ore for this much that’s the p
that’s the price um given the units x number of units dollars per ton right
so find the revenue and the marginal revenue functions
and then answer this question here alright so let’s find the revenue function

01:21
so the revenue is r of x which is going to be x times p so the price
and then times how many of them were sold now they give us an expression for p
so we can write it as a function of x so this will be x times 225 minus 0.25 x
and if we just simplify this a little bit so this will be 225 x
and then minus 0.25 x squared so now the marginal revenue [Music]
so the marginal revenue is r prime which is 225
minus and then now multiply that by two so zero point five

01:22
x or another way to write that is one half times 450 x minus x
okay so there’s the revenue and there’s the marginal revenue
and i write it like this so we can see exactly where the marginal revenue is
zero in case they want to ask that but we found the first part find the revenue
and the marginal revenue now it asks us at what level of
production would the company obtain the maximum revenue so this is an
optimization problem we’re trying to maximize revenue
now if we look at this function here it is continuous and differentiable
everywhere it’s just a parabola and we’re looking at
you know where x is 450 that’s where the derivative
is zero so we can look less than 450 we can look at equal to 450 and we can

01:23
look greater than 450 and we’ll look at our derivative and
we’ll make a conclusion so when we’re less than 450 for example
now of course x is always positive because x is the number of units being
you know x is the number of uh tons of ore right so x is greater than
or equal to zero but anyways we choose something like one like one ton
right so that’ll be positive and if we choose something like 500 tons
then this will be negative so it’s increasing and then it’s
decreasing and so this will be a relative max at 450. now
greater than 450 it’s still going to keep going down
so this relative max actually has to be an absolute max

01:24
so we have an absolute max at x equals 450 so what level of production would the
company obtain the maximum revenue so 450 units is the correct answer there
absolute max 450 okay so or here we could say we know the units 450 tons
there we go all right let’s look at another example let’s let’s
optimize some profit this time okay so let’s see what we got here
a fertilizer producer finds that he can sell its product
at a price so there’s the price over the demand function as we called it earlier

01:25
and we want to know all right so the total total production cost is given to us
that’s how they [Music] compute the cost of producing a unit
and we’re told that the production capacity
is at most 1000 that’s how many units they can make they can’t make any more
and then how many units must be manufactured and sold
in that time to maximize the profit okay so let’s start off by looking for
the revenue first so the revenue is r of x
which is as we said before it’s x times the
the price the number of the number of them sold
times the price of them and they give that to us as a 300 minus 0.1 x

01:26
and so this is the revenue function or if you want to write it like this
300 x minus 1 over 10 x squared and now but we’re looking for profit here
um we’re looking to maximize the profit right so
the profit p of x is the revenue minus the cost
and we just found the revenue function so it’s 300 x
minus 1 10 x squared so revenue minus the cost and they give us the cost
function here so it’s 15 000 plus 125 x plus 0.0025
i’ll write it down here plus 0.0025 x squared all right so there’s the

01:27
revenue minus the cost but you know can we make this look nicer
so let’s go with here with minus 15 000 plus 175 x putting those together
and then minus and then putting those together we’re going to get
0.125 x squared and so this is the profit right here
now the x is bound because they specifically told us at most 1 000 units so
i’m going to say here zero to one thousand so this is our profit function
p of x equals so i think we can fit all that right there
so there’s the profit on the closed interval zero to one thousand

01:28
right we we have to make a positive number of them
and then no more than 1 000 units in a specified time whatever that time
period is all right so i have a continuous function
on a closed interval so i’m going to use the extreme value theorem
so i’m going to look at the profit on the zero
on the boundary on the zero and 1000 but also
you know do we have any critical points so let’s find the derivative
so the derivative is 175 minus 0.25 0 x and that is going to be 0
so x equals 175 divided by 0.25 and that if you calculate that out you

01:29
get 700 units so there’s the derivative we set it equal to zero
we find our critical number it turns out to be seven
so now to do use the extreme value theorem i’m going to use p of 0 and p of 1000
those the boundaries and then the critical number p of 700 and we will find our
maximum profit here so p of zero so where’s p p of zero right here is
minus fifteen thousand that one’s easy p of one thousand so now i gotta go put
a thousand in here and a thousand in here and when i do all that
i get forty eight thousand five hundred and now i have to go put 700 in here and
700 in here and calculate all that up and here i get 59 750.

01:30
so we can see right here this is the max profit
the max profit it happens when 700 units are sold
so how many units must be manufactured 700
sold in that time period to maximize the profit
and there is the maximum profit so by the extreme value theorem we only
needed to check the boundary and the critical numbers because we have
a continuous function on a closed interval and so we use the
derivative to find the critical number and then we tested that there also
all right so very good so here on this problem we started with the revenue
and we generated the profit function because they gave us the cost so notice
in this problem they they gave us information to find the
revenue they gave us the cost putting those together we’re able to
find the profit and then we’re able to maximize the

01:31
profit and we did so using the extreme value theorem
all right very good next example all right so this one looks kind of long
and let’s see here we have a firm determines that x
units of its product can be sold daily at p
dollars per unit and we’re given here the cost 3000 plus 200x
so first we want to find the revenue function okay we’ve done that
then we want to find the profit function okay we do profit is
revenue minus cost all right so let’s let’s do a and b here
here we go let’s do a first so the revenue is

01:32
the revenue is p times x the number of units sold times the price of them
so p of x well that looks like multiplication so i’ll just say x times p
so x times p okay so we need to go solve for the p over here don’t we so
you know switch x’s and p’s so it’s 1 000 minus x
all right and so said differently this revenue is 1000 x minus the x squared
and that is the revenue function right there so let’s identify that as revenue
all right revenue okay so part b is going to be um find the profit

01:33
so profit will be revenue minus cost and so we’re going to look at this one
minus that c i think we can squeeze it in so here’s revenue
minus the cost which is 3 000 plus 20 x all right so there’s profit revenue
minus cost and let’s see if we can write that better
so we’re going to get minus 3 000 plus 980 right put those together
minus x squared so this will be the profit function right there this
put profit function profit there’s a and b all right so now let’s look at c

01:34
okay so assuming the production capacity
is at most 500 units per day so that’s a boundary condition there
determine how many units the company must produce
and sell each day to maximize the profit so i want to maximize the profit here
and we want to so i’ll put here max p of x um subject to
zero of course you the least you can do is zero for the number
of units and then it said 500 units so that allows us to use the extreme
value theorem so we can use extreme value theorem let’s find p prime so
p prime here is the 980 minus the 2x set that equal to zero

01:35
so the critical number is x equals 490 and so that’s going to give us the
maximum we think well part d says find the maximum profit
let’s go ahead and make sure that this is the this is the number of units that
gives us so let’s look at p of zero p of 500 so those are the boundary
um now also make sure that’s in the boundary right that’s in here
otherwise you would discard it so 490 let’s check that one also so we’re
checking the boundary and the critical numbers into
our profit function here so this one is minus 3 000 and this one
500 and we put it 500 here and 500 here we crank that out and we’re going to get

01:36
237 000 and now we take our 490 and put it here and here
we crank that out oops not mark it out crank it out and we’re going to get 237
000 one hundred dollars and so this right here will give us the max profit
that occurs when the number of units is 490.
when you produce 490 units don’t go to 500
you won’t make it’s quite as much money okay so now the last part is
alright so that was d now for e at what price per unit must be charged
to obtain the maximum profit so what price per unit
so we’re going to use x is 490 that’s how many units
and then if we go back up in here and look at our p

01:37
p so that x is telling us that 1000 minus p so 490 is the x equals to 1
000 oh 1 000 minus p so solving for p to find the price
and then we’re just going to say p is equal to 1000 minus the 490 so the price
is going to be 510 per unit units so the price so what price per unit must
be charged to obtain maximum profit so we want to charge 510 dollars
we want to produce 490 of them and this is the profit we would get and
it would be the maximum profit all right this is a lot of fun so in that case

01:38
let’s do one more so here we go this is our last one it looks like from
that right there so here we go the chemical manufacturer sells
sulfuric acid in bulk and there’s the price
per unit and the daily total production cost
is given by that expression there and the daily production capacity is at most
seven thousand so how many units of acid must be manufactured
and sold daily to maximize profits so that’s all part a
okay so we know profit is equal to revenue minus cost
and let’s see what we can get for this right here

01:39
so the revenue here is uh 100 per unit so 100 times x right so 100
times how many of them we produce the number of units
minus and now the cost so the cost is given right here
so i’m going to have minus cost so i’m going to multiply that cost by
through by a negative sign so minus 1000 [Music] oops so um
i think there might be an extraneous zero in there this is 100 000
and then minus 50x and then minus 0.0025 x squared so there’s the profit right

01:40
there it’s the revenue minus the cost so i just multiplied the whole cost
minus the whole cost so the cost now has a bunch of negatives in it and so we
want to um think of this profit function with restriction of 7 000 units so
on 0 to 7 000 7 thousand yeah seven thousand seven thousand units
are on this interval on this closed interval so we have a continuous function
on a closed interval and you know we could put that together and just say 50x
all right let’s do that fine so p of x is the minus 100 000
plus 50x and then minus the 0.0025 x squared and then now we can say over here

01:41
on now i don’t have enough room so on 0 to 700. seven thousand
all right anyways um how many units must be manufactured to maximize
profit so i’m going to use extreme value theorem why because i have a continuous
function on a closed interval so extreme value theorem says
the function must have an absolute maximum and must have an absolute minimum
we’re only interested in the maximum for this problem so we need to evaluate p
on the boundary and any critical numbers so to find the critical numbers we need
to find the derivative so the derivative is we got the 50 and then minus 0.005 x
right multiplied by 2. and so x is 50 divided by .005 and that’s going to be

01:42
10 000. so ten thousand units right but actually is not in the interval there so
this this is not going to be part of the
solution at the end um if you are trying to maximize profits
they for some reason they can only make 7 000 units
so if you could make 10 000 units you would make more profit but
you know when we look at um when we look at part a here how many units
to to to maximize the profit you know to to do the
extreme value theorem we need to check the boundary and the
critical numbers but the critical numbers not in here
so the the maximum is going to come from the boundary and this will be 7 000.

01:43
so um at zero it’s just going to be minus right we have zero and a zero so
we have a minus one hundred thousand well that’s definitely not their maximum
profit what about for seven thousand now if we plug in seven thousand here
seven thousand here and say hey if we sell
seven thousand units our maximum number of units
possible and then we’ll calculate the profit so we plug in seven thousand
and seven thousand we crank out that number and that number is one hundred
and twenty seven thousand and five hundred and that’s the max profit
if they make seven thousand units they get the maximum profit
and that according to this information over here their manufacturing plant
for whatever reasoning their manufacturing process
can only build seven thousand um a day right so there’s part a

01:44
now part b part b said would it benefit the manufacturer to expend
to expand the daily production and the answer to that is
yes they want to expand it past 10 000 in fact they actually just want to
expand it to 7 000 to 10 000 so yes yes expand manufacturing to 10 000 units
well wait a minute we didn’t actually calculate the profit what is p
of 10 000 we know what we get when we produce 7 000 units we get that
amount of profit what happens when we did 10 000 units
so you know calculate that number um and then you’ll
you’ll get the the maximum there you’ll get maximum profit now
how do i know that well you can plug in the 10 000

01:45
and crank that number away because we’re using the extreme value theorem
and you’ll get more profit now if you didn’t use
extreme value theorem if for some reason you wanted to use
the first derivative test then you could look
less than 10 000 and greater than 10 000 and realize that this is going to be an
absolute max um but the answer to the question with a benefit yes
this is going to be the max profit right here and you simply find that number by
plugging in the 10 000 here and here and let’s see here okay so part c
use marginal analysis to approximate the effect

01:46
on profit if daily production could be increased by one additional unit so
in other words the profit of seven thousand and one minus the
profit of seven thousand which is the uh profit from one additional unit
that can be approximated by the derivative at seven thousand
so this is our linear approximation formula so remember this is the profit from
one additional unit right that’s the profit of seven
thousand and one minus the profit of seven thousand
7000 and so we can approximate it by plugging into the derivative here um
and and getting so here’s the derivative right here
and we can plug in 7000 into there so let’s do that over here

01:47
so this will be the derivative at 7 000. this will be 50 minus 0.005
right so there’s the derivative right there times the x which is the 7 000.
so there’s the marginal profit we found it by calculating that right
there and when you calculate that you get 15. and this is the marginal profit
[Music] and so it asks us to use marginal analysis to approximate
so this is approximately here the a profit from one additional unit
past seven thousand is approximately fifteen dollars that’s the marginal
profits at seven thousand so it’s approximately fifteen dollars um

01:48
price per unit yeah so fifteen dollars [Music] and that’s part part c
okay so there we go we used the marginal analysis there by approximating
um and so that solves this example and we’re done with all the examples
so now let’s see what’s next [Music] okay so now we’re going to look at some
exercises and there’s going to be some exercises
here all these exercises are going to be optimization problems
and the idea is to look at them and figure out
which ones you can solve and which ones you can’t and so let’s go here
and look at these problems here sorry so

01:49
um yeah follow the procedure that i just went over make sure you draw a picture
for each one make sure you label the variables and those are the first three
and here’s four and five um and some of these problems
now some of these problems you will not be able to solve just
like within five minutes you’ll want to think about it and try to come up with
ideas um and so there’s six and seven and for the last batch we have eight and
nine and so there’s nine problems right there for you to practice on
now if you get stuck on any of them or if you just want to compare your work
and uh yeah if you’re interested in these exercises let me know
okay so um now i want to say thank you for watching and um you know if you
enjoyed this video please like or subscribe below

01:50
and up next we have the la hoppitals rule and indeterminate forms oh i love that
episode i love those limits we’re going to calculate limits
in the indeterminate forms like 0 0 or infinity over infinity and we’re
going to use derivatives to calculate those
limits so yeah we use limits to define derivatives
now we’re going to use derivatives to help us find limits
and um so i can’t wait to see you on that episode there
and uh yeah if you um want to find me on social media the links are
below and i look forward to seeing you next time thanks for watching
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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