Main content

## Calculus 1

### Course: Calculus 1 > Unit 5

Lesson 11: Solving optimization problems- Optimization: sum of squares
- Optimization: box volume (Part 1)
- Optimization: box volume (Part 2)
- Optimization: profit
- Optimization: cost of materials
- Optimization: area of triangle & square (Part 1)
- Optimization: area of triangle & square (Part 2)
- Optimization problem: extreme normaline to y=x²
- Optimization
- Motion problems: finding the maximum acceleration

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Optimization: box volume (Part 2)

Finishing up the last video by working through the formulas. Created by Sal Khan.

## Want to join the conversation?

- 2:45But what are the odds that "12x^2 - 200x + 600" just happens to be in the correct format to use it in the quadratic formula? What if it were: "-200x + 600 = 0" or "12x^3 - 200x + 600 = 0"?(30 votes)
- Hi Steven. That is an interesting question. Actually it happens to be a quadratic by chance, or more precisely, because the people who invented that exercise wanted the solver to combine his knowledge in calculus and the quadratic formula.

In many other cases it is not so simple to get an exact answer, and you would have to use the "graphing method" Sal explained or make some numerical approximations (either to maximize the function or to set the derivative equal to zero).(60 votes)

- Why did you use 20-2x to find the domain and not 30-2x?(22 votes)
- Its because the smallest side is 20", and when x=10" the smallest side is folded and volume=zero.(4 votes)

- I have an issue with a simular problem. My sheet of metal is given as 40cm x 80cm, and when I follow this method, I end up trying to square root a negative number. Which of course isn't possible, at least without moving into the realm of complex numbers which won't help with my problem.(6 votes)
- Instead of doing the second derivative test, couldn't you have just plugged in the x values into the original equation?(5 votes)
- You can do it that way, but it requires more points to determine the concavity. In most cases, the second derivative test is the easiest way to do it.(22 votes)

- So, what units would this problem be answered in? Would it be unitless, or would you say "units cubed?"(4 votes)
- It would be inches^3 as the original measurements of the cardboard were given in inches(15 votes)

- How does Khan simplify 4^3 - 100x^2 + 600x to 12x^2 - 200x + 600? I am very new to calculus and just came to watch these videos for optimization, so please explain very clearly.(5 votes)
- Try searching Khan Academy for 'power rule'.(8 votes)

- I'm having problems understanding what I'm looking for. I don't quite fully understand the concept of optimization in total. I have a test today in math and I'm scared I'm not going to do well. Could you please help me understand this? Thank you :)(2 votes)
- I don’t know how much help you can get from a brief reply here, but I’ll offer some comments for what they’re worth.

Optimization problems have to do with finding a tipping point. Something is getting better up to a point, and then it starts to get worse. It’s getting bigger, then it starts to get smaller. Or it’s getting smaller, then it starts to get bigger.

We find those tipping points by looking at the derivative, which is the rate at which something is changing. As long as the rate of change is in the “good” direction (which may be up or down, depending on what you’re optimizing), we keep going. The tipping point is where the derivative starts to go in the other direction. It’s the top of a frown-shaped curve or the bottom of a smile-shaped curve. The rate of change at these exact moments is zero, so we hunt for optimization points by finding the derivative and then determining where the derivative is equal to zero (the “critical points”).

You need to be aware that some functions will have two or more critical points, and you have to use other tests to determine which one is optimal in terms of the way the problem was set forth. Some functions have critical points that aren’t turning points (for example, y = x^3 has a critical point at x = 0 but doesn’t turn down). And some teachers want to test your understanding by giving you a problem where there’s only one critical point, which is a true turning point, but because of the way the problem was stated, this is a worst solution instead of an optimal one.

Good luck!(9 votes)

- how would you find the dimensions of an open top box with a minimum surface area and a volume of 32000 cubic cm, I'm guessing partial derivatives but lost other than that(2 votes)
- In theory you have three variables to work with: height, width and depth. In practice, it's fairly obvious that the box has to be square, so you have only two variables, one for the height of the box and one for the length of a side (width and depth). If you're expected to prove this part, you begin by showing that the rectangle with minimum perimeter for a fixed area is a square.

With the problem reduced to height and length of a side, you can create a formula for the surface area (remembering to include the bottom of the box) and set it equal to 32,000. Differentiate and find the place where the derivative is equal to zero, then confirm that this is a minimum.(7 votes)

- In maxima-minima word problems, if something is not given as a constant, and is not mentioned at all, can we take it to be variable?(4 votes)
- This question is tough to answer because it is not clear exactly what kind of situation you're in.

The word problem could be a general problem, meaning that you should solve the problem for any imaginable number, which gives you an answer, in this case a formula for solving that type of problem.

If it's a specific problem, you can see the previous video on the graphical solution to find out how Sal managed to find an expression for the different lengths:

Basically - if you want to find an exact solution you need to be given enough variables to be able to compute the exact answer in the end.

That does not mean these things are written down for you, it might be implicit in the problem itself, and the problem might be constructed for you to be able to deduce the necessary equation that allows you to find the maximum/minimum of the function

Hope that helps!(2 votes)

- isnt x also not equal to 15 because of the polynomial (30-2x)?

If x = 15 then (30-2x)=0 and then V(x) will also be equal to 0(3 votes)- You are right, but 𝑥 ≠ 15 is already included in 0 ≤ 𝑥 ≤ 10.(4 votes)

## Video transcript

In the last video, we were
able to get a pretty good sense about how large of an x we
should cut out of each corner in order to maximize our volume. And we did this graphically. What I want to do
in this video is use some of our calculus
tools to see if we can come up with the same or maybe
even a better result. So to do that, I'm going
to have to figure out the critical points of our
volume as a function of x. And to do that, I need to take
the derivative of the volume. So let me do that. And before I even do that,
it'll simplify things so I don't have to use the
product rule in some way and then have to
simplify that, let me just multiply
this expression out. So let's rewrite volume as a
function of x is equal to-- and I'll write it all in yellow. So it's going to
be x times-- I'll multiply these two
binomials first. So 20 times 30 is 600. Then I have 20 times negative
2x, which is negative 40x. Then I have negative 2x times
30, which is negative 60x. And then I have negative
2x times negative 2x, which is positive 4x squared. So this part over
here simplifies and I can change the order to
4x squared minus 100x plus 600. I just switched the order
in which I'm writing them. So that's that. And so I can rewrite
the volume of x as being equal to x times all
of this business, which is-- let me make sure I
have enough space, let me do it a
little higher-- which is equal to 4x to the third
power minus 100x squared plus 600x. And that'll be pretty
straightforward to take the derivative. So let's say that v prime of
x is going to be equal to-- I just have used the power rule
multiple times-- so 4 times 3 is 12. x to the 3 minus 1 power
12x squared minus 200 times x to the first power,
which is just x, plus 600. And so now we just
have to figure out when this is equal to 0. So we have to figure
out when 12x squared minus 200x plus
600 is equal to 0. What x values gets my
derivative to be equal to 0? When is my slope equal to 0? I could also look
for critical points where the derivative
is undefined, but this derivative
is defined especially throughout my domain of x that
I care about, between 0 and 10. So I could try to
factor this or try to simplify this a little bit. But I'm just going
to cut to the chase and try to use the
quadratic formula here. So this is, the x's
that satisfy this is going to be x is going to
be equal to-- So negative b. So it's 200. Negative, negative
200 is positive 200. Plus or minus the square
root of b squared, which is negative 200 squared. So I could just write
that as 200 squared. Doesn't matter if it's negative
200 squared or 200 squared, I'm going to get the same value. So let me give myself
some more space. So negative 200 squared. Well, that's going
to be 4 with 4 0's. 1, 2, 3, 4. So that's going to
be 40,000 minus 4ac. So minus 4 times
12, 12 times 600-- I still didn't give myself
enough space-- times 600. All of that over 2 times a. So all of that over 24. And I'll take out the calculator
again to try to calculate this. So let me get out
of graphing mode. All right, so first I'll
try when I add the radical. So I'm going to get 200 plus
the square root of 40,000. I could have just written
that as 200 squared, but that's fine. 40,000 minus 4
times 12 times 600. And I get 305, which I
then need to divide by 24. Which I'll divide by
24, and I get 12.74. So one of my possible x's. So it equals 12.74. And now let me do
the situation where I subtract what I had
in the radical sign. So let me get my
calculator back. And so now let me
do 200-- I probably could have done this
slightly more efficiently, but this is fine-- minus the
square root of 40,000-- 1, 2, 3-- minus 4 times 12 times 600. And I get that's
just the numerator. And then I'm going to divide
that by 24, and I get 3.92. Did I do that right? 200 minus 40,000 minus 4 times
12 times 600, all of that divided by 24. My previous answer divided
by 24 gives me 3.92. So it's 12.74 or 3.92. Now which of these can I use? Well x equals 12.74 is outside
of our valid values for x. If x was equal to
12.74, the x's would start to overlap
with each other. So x cannot be 12.74. So we get a critical point
at x is equal to 3.92. And you could look at the
graph and you could say, oh, well look, that looks
like a maximum value. But if you didn't have the
graph at your disposal, you can then do the
second derivative test and say, hey, are we concave
upwards or concave downwards when x is equal to 3.92? Well, in order to do the
second derivative test, you have to figure out
the second derivative is. So let's do that.
b prime prime of x is going to be equal to 24x,
24 times x to the first, minus 200. And you can just
look at inspection that this number right
over here is less than 4. So this thing right over here
is going to be less than 100. You subtract 200. So we can write the
second derivative at 3.92 is going to be less than 0. You can figure out what the
exact value is if you like. So because this is less than
0, we are concave downwards. Another way of saying
it is the slope is decreasing the entire time. Concave downwards. When the slope is
decreasing the entire time, our shape looks like that. The slope could start off
high, lower, lower, gets to 0, even lower, lower, lower. And we even saw that on
the graph right over here. And since it's
concave downwards, that implies that our
critical point that sits where the interval
is concave downward, that critical point
is a local maximum. So this is the x value
at which our function attains our maximum. Now what is that maximum value? Well, we could type that back
in into our original expression for volume to
figure what that is. So let's figure out what the
volume when we get to 3.92 is equal to. What is our maximum volume? So get the calculator back out. It's obviously roughly 3.92. I could use this exact value. Actually, I'll just use 3.92
to get a rough sense of what our maximum value is,
our maximum volume. So it'll be 3.92. I'll just use this
expression for the volume as a function of x. 3.92 times 20 minus 2 times
3.92 times 30 minus 2 times 3.92 gives us-- and we deserve a drum
roll now-- gives us 1,056.3. So 1,056.3, which
is a higher volume then we got when we just
inspected it graphically. We probably could have gotten
a little bit more precise if we zoomed in some and then we
would have gotten a little bit better of an answer,
but there you have it. Analytically, we were
able to actually get an even better
answer than we were able to do at least on that
first pass graphically.