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### Course: Differential Calculus>Unit 5

Lesson 11: Solving optimization problems

# Optimization: box volume (Part 1)

If you are making a box out of a flat piece of cardboard, how do you maximize the volume of that box? Created by Sal Khan.

## Want to join the conversation?

• How to approach the optimization questions in calculus ? Each question seems to be very different in its approach . Can someone please summarise the steps I have to do in it .
• A quick guide for optimization, may not work for all problems but should get you through most:
1) Find the equation, say f(x), in terms of one variable, say x.
2) Find the derivative of that function.
3) Find the critical points of the derivative where f'(x)=0 or is undefined
4) Test those points with the second derivative to determine how they look on a curve, whether it is concave up or down, and whether the point would then be a minimum or maximum.

That point, x, will be the optimized (min or max) value for x in your initial equation, and from there you can determine how to give the answer based on what's asked of you.

I hope this helps a bit! c:
• why x is greater or EQUAL to zero? if we assume x is zero then the whole volume would become zero right? is that possible?
• It is possible to have 0 volume. In mathematics, a plane is a flat, two-dimensional surface. A plane is the two dimensional analogue of a point (zero-dimensions), a line (one-dimension) and a solid (three-dimensions).
• Why doesn't Sal use 30-2x in his range 0<x<10? Why does he only use x and 20-2x?
• This is because if we solve 30-2x is greater than or equal to x, then we get x is less than or equal to 15. But since x has to be less than or equal to 10 to make the (20-2x) dimension positive, it overrules the condition from the 30-2x. This makes sense in cases where you may have an x value greater than 10 but less than 15, such as 13, which would make the (30-2x) dimension positive, but it would make the (20-2x) dimension negative. Therefore, (x is less than or equal to 10) rules out the condition of x being less than or equal to 15 in order to keep the (30-2x) and (20-2x) dimensions positive, preventing a negative volume.
• At : 30 - 2x is the width, isn't it?
• The width is 20 - 2x, the depth is 30 - 2x. But it doesn't really matter in this case, because if you turned the box 90 degrees, the depth would become 20 - 2x and the width would become 30 - 2x.
• At , how can x be equal to zero ?
• At this point, Sal is merely eliminating as impossible the case where x is negative, so he states that x ≥ 0. A minute or so later in the video he points out that x = 0 is also impossible, and eliminates that case.
• Wouldn't the volume simply get larger the smaller x is?
• If x was really small, like 1/1000 of an inch, you would only be folding the edges of the box up 1/1000 of an inch. So you'd get a very wide, shallow box. The area of the bottom would be very close to 600, but then we multiply that by the height of 1/1000 to get a volume of 0.6 cubic inches.

So for very large x and very small x, we get very small volumes.
• But wouldn't larger X values simply make the function x(20-2x)(30-2x) larger? 500(20-1000)(30-1000)= 475300000.
• That's true, but in this case, we have a domain to x- we don't have 500 units of space to fold on our sheet of cardboard.
• How were we able to determine 0 <= x =< 10. More specifically, how was 10 determined?
(1 vote)
• We're cutting x*x squares out of each corner, so x can't possibly be more than half of the width of the piece of cardboard, which is 20.