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## Solving optimization problems

Current time:0:00Total duration:12:40

# Optimization: cost of materials

AP Calc: FUN‑4 (EU), FUN‑4.B (LO), FUN‑4.B.1 (EK), FUN‑4.C (LO), FUN‑4.C.1 (EK)

## Video transcript

A rectangular storage
container with an open top needs to have a volume
of 10 cubic meters. The length of its base
is twice the width. Material for the base
costs $10 per square meter. Material for the sides
costs $6 per square meter. Find the cost of the material
for the cheapest container. So let's draw this
open storage container, this open rectangular
storage container. So it's going to
have an open top. So let me draw its open
top as good as I can. So it's going to
have an open top. That's the top of my container. And then let me draw the sides. Just like that. So it might look
something like that. And then I could draw--
and since it's open top, I can see through, I could see
the inside of the container as well. So the container would
look something like that. And so what do they tell us? They tell us that the volume
needs to be 10 cubic meters. So let me write that down. The volume needs to be
equal to 10 meters cubed. The length of its base
is twice the width. So the length, let's
call the width x, so the length is going
to be twice that. It's going to be 2x. That's what they tell
us right over here. They tell us the
material for the base costs $10 per square meter. So this area right
over here-- [INAUDIBLE] if I was transparent I could
continue to draw it down here. But this right over
here, that material costs $10 per square meter. Let me label that
$10 per square meter. And then they say
material for the sides costs $6 per square meter. So the material over here
costs $6 per meter squared. So let's see if we can
come up with a value or how much this box would cost
to make as a function of x. But x only gives us the
dimensions of the base. We also need a
dimension for height. So it'll be a function
of x and height for now. So let's write h as the
height right over here. So what is the cost of
this container going to be? So the cost is going to be
equal to the cost of the base. Well, the cost of the base
is going to be $10 times-- I'll just write 10. This is going to be 10
times the area of the base. Well, what's the
area of the base? Well, it's going to be the
width times the length. So 10 times x times 2x. That is the cost of base. And now what's going to
be the cost of the sides? Well, the different
sides are going to have different dimensions. You have this side
over here and this side over here, which have
the same dimension. They both have an
area of x times h. You have x times h. And then our material
is $6 per square meter. So it's 6 times x times h
would be the cost of one of these side panels. So for two of them we
have to multiply by 2. So plus 2 times 6 times h. And then we have
these two side panels. We have this side
panel right over here and we have this side
panel right over here. The area of each of them
is going to be 2x times h. So it's going to be 2x times h. The cost of the material
is going to be 6. So the cost of one
of the panels is going to be $6 per square
meters times 2xh meters squared. But we have two of these panels. One panel and two panels. So we have to multiply by 2. And so we will get-- so
this is right over here, this is the cost of the sides. And so let's see if
we can simplify this. And I'll write it all
in a neutral color. So this is going
to be equal to 10. Let's see. 10 times 2 is 20. x times x is x squared. And then you have
2 times 6 times xh. So this is going
to be plus 12xh. And then this is going
to be 2 times 6, which is 12 times 2 is 24xh plus 24xh. So this is going to be equal
to 20x squared plus 36xh. So this is going to be my cost. But I'm not ready
to optimize it yet. We don't know how to optimize
with respect to two variables. We only know how to optimize
with respect to one variable, and maybe I'll say let's
optimize with respect to x. But if we want to optimize
with respect to x, we have to express h
as a function of x. So how can we do that? How can we express h
as a function of x? Well, we know that the volume
has to be 10 cubic meters. So we know that x, the width
times the length times 2x times the height times h
needs to be equal to 10. Or another way of saying
that, this tells us that 2x squared h,
2x squared times h needs to be equal to 10. And so if we want h
as a function of x, we just divide both
sides by 2x squared. And we get h is equal
to 10 over 2x squared. Or we could say that h is
equal to 5 over x squared. And then we can substitute
back right over here. h is equal to 5 over x squared. So all of this business is
going to be equal to 20 times x squared plus 36 times
x times 5 over x squared. So our cost as a function
of x is going to be 20x squared 36 times 5. Let's see, 30
times 5 is 150 plus another 30 is going to be 180. So it's going to be plus
180 times, let's see, x times x to the negative
2, 180x to the negative x to the negative 1 power. So we finally have cost
as a function of x. Now we're ready to optimize. To optimize, we just
have to figure out what are the
critical points here and whether those
critical points are a minimum or a maximum value. So let's see what we can do. So to find a critical point,
we take the derivative, figure out where the derivative
is undefined or equal to 0, and those are our
candidate critical points. And then from the
critical points we find, they might be
minimum or maximum values. So the derivative of c of
our cost with respect to x is going to be equal to 40
times x minus 180 times x to the negative 2 power. Now, this seems-- well
it's defined for all x except for x equaling 0. But x equaling 0 is
not interesting to us as a critical point
because then we're going to have a degenerate. This is going to
have no base at all. So we don't want to worry
about that critical point. We would have no volume at
all, so it would not work out. And actually, if x
equals 0 then our height is undefined as well. So this was defined
for everything else, for anything other
than x equals 0. So let's see when
this derivative is equal to 0 in our search for
our potential critical points. So when does-- I'll
do it over here. When does 40x minus 180x
to the negative 2 equal 0? Well, we could add the 180x to
the negative 2 to both sides. We get 40x is equal to 180. And I could write it
as 180 over x squared. Now let's see. We could multiply both sides
of this equation by x squared and we would get 40x to
the third is equal to 180. Divide both sides by 40. You get x to the third
is equal to 180 over 40, which is the same
thing as 18 over 4, which is the same
thing as 9 over 2. And so if we want
to solve for x, we get that x is equal
to a critical point. We get a critical point of x is
equal to 9/2 to the 1/3 power, the cube root of 9/2. So let's see. Let's get an approximate
value for what that is. So if we take 9/2,
9 divided by 2-- I guess you could
call that 4.5-- and we want to raise
it to the 1/3 power. To the 1/3 power we get 1.65. So it's approximately equal
to 1.65 as our critical point. Now the way the
problem is asked, we're only getting one
legitimate critical point here. So that's probably going
to be the x at which we achieve a minimum value. But let's use our second
derivative test just in case to make sure that
we're definitely concave upwards over here,
in which case, this will definitely be the
x value at which we achieve a minimum value. So the second derivative. I'll do it right over here. The second derivative
of our cost function is just the derivative
of this, which is going to be equal to 40 minus
180 times negative 2, which is negative 360. So it's going to be plus
360 over x to the 1/3. The derivative of this is
negative 2 times negative 180, which is positive 360x to
the negative 3 power, which is exactly this right over here. So when x is equal to 1.65,
this is going to be positive. This is going to be positive. So let me write this down.
c prime prime of 1.65 is definitely greater than 0. So we're definitely concave
upwards when x is 1.65. Concave upwards, which
means that our graph is going to look
something like this. And so where the
derivative equal to 0, which is right over there,
we are at a minimum point. We are minimizing our cost. And so if we go back to the
question, the only thing that we have to do now--
We know the x value that minimizes our cost. We now have to find the cost of
the material for the cheapest container. So we just have to figure
out what our cost is. And we already
know what our cost is as a function
of x, so we just have to put 1.65
into this equation. Evaluate the function at 1.65. So let's do that. Our cost is going to be
equal to 20 times 1.65. I should say
approximately equal to, because I'm using
an approximation of this original value. 1.65 squared plus 180. I could say divided by 1.65. That's the same
thing as multiplying by 1.65 to the negative 1. So divided by 1.65,
which is equal to 163. I'll just say $163.5. So it's approximately. So the cost-- let me
do this in a new color. We deserve a drum roll now. The cost when x is 1.65 is
approximately equal to $163.54. So $163.54, which is
quite an expensive box. So this is kind of expensive. This is fairly
expensive material here. Although it's a
fairly large box. 1.65 meters in
width, and it's going to be twice that in length. And then you could figure out
what its height is going to be. Although it's not
going to be too tall. 5 divided by 1.65 squared. I don't know, it'll be roughly
a little under two meters tall. So it actually is quite
a large box made out of quite expensive material. The minimum cost to make this
box is going to be $163.54.

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