# Calc One Maximization Problem

• Oct 6th 2007, 12:55 PM
topsquark
Calc One Maximization Problem
I recently completed a Calculus proficiency exam on tutor.com. Here is a question they asked me:

An open box is to be made from a 20 inch square piece of cardboard by cutting four equal square pieces from each corner and turning up the sides. What is the side length of the pieces that should be cut out so that the box will have maximum volume?

This question can be done, using partial derivatives, but this solution (that the cardboard is a square with sides $\sqrt{20}$ inches long) was not listed as one of the possible answers. Not to mention the exam was supposed to use Calc I techniques, of which I doubt partial derivatives are included.

Is there any way to generate a solution here without resorting to partial derivatives? (I finally answered the question by process of elimination: only one answer listed was physically possible.)

-Dan
• Oct 6th 2007, 01:56 PM
galactus
Hey TP:

If you let the side of one of the little squares be x, then the volume of the box would be:

$V=x(20-2x)(20-2x)=4x^{3}-80x^{2}+400x$

Differentiate: $12x^{2}-160x+400=3x^{2}-40x+100=(x-10)(3x-10)$

The solutons are 10/3 and 10

x=10/3 is the one to use.
• Oct 6th 2007, 02:03 PM
topsquark
Quote:

Originally Posted by topsquark
An open box is to be made from a 20 inch square piece of cardboard by cutting four equal square pieces from each corner and turning up the sides. What is the side length of the pieces that should be cut out so that the box will have maximum volume?

Quote:

Originally Posted by galactus
Hey TP:

If you let the side of one of the little squares be x, then the volume of the box would be:

$V=x(20-2x)(20-2x)=4x^{3}-80x^{2}+400x$

Differentiate: $12x^{2}-160x+400=3x^{2}-40x+100=(x-10)(3x-10)$

The solutons are 10/3 and 10

x=10/3 is the one to use.

(Doh) Ah, I see what happened. There are two possible ways to read the phrase "20 inch square piece of cardboard." You took it to be a 20 in x 20 in piece of material (which generates the only possible physically correct answer) and I took it to mean that the area of the cardboard is 20 in^2.

Hmmm... Anyone else think the wording here is ambiguous? Or am I just having a perception problem today? (Drunk)(Drink)

-Dan
• Oct 6th 2007, 02:30 PM
ticbol
20 inch square is not 20 square inches.

Umm, the in^2.
20 in^2
No, it's not that here.
Otherwise it is written in in^2.

20 inch square is more like 20-inch square.
• Oct 6th 2007, 02:44 PM
galactus
Quote:

I recently completed a Calculus proficiency exam on tutor.com.
I done that a few months back. I passed all the tests with flying colors, but am unable to participate because I am relegated to a damn, dial-up.(Swear)
• Oct 6th 2007, 02:50 PM
topsquark
Quote:

Originally Posted by galactus
I done that a few months back. I passed all the tests with flying colors, but am unable to participate because I am relegated to a damn, dial-up.(Swear)

Sorry to hear that. My woes include moving (for the second time in a month) and from what I've been told I'll lose my DSL for about two weeks. Ugh.

-Dan