Optimization (Using 1st/2nd Derivative Tests): making a box by cutting out corners...

[crybloodwing]

From a piece of cardboard 30in by 30in, square corners are cut out so the sides can be folded to make a rectangular box. What dimensions of the squares will yield a box of maximum volume?

1. Drew a picture of what it would look like an labeled it.
2. Wrote out two formulas. The volume formula and the formula for one side.

To get the maximum volume, I need to find the 1st derivative of the volume equation.

However, doing this, I get a critical number that is negative and also not the right answer.

The answer to this problem should be 20"x20"x5". 20 for length/width, 5 for the height.

I am not sure what I am doing wrong?

 

[JeffM]

You started well.

\(\displaystyle V(x) = y^2 * x \text { and } y = 30 - 2x \implies \)

\(\displaystyle V(x) = (30 - 2x)^2 * x = (900 - 120x + 4x^2)x = 4x^3 - 120x^2 + 900x \implies\)

\(\displaystyle V'(x) = WHAT?\)

 

[crybloodwing]

You started well.

\(\displaystyle V'(x) = WHAT?\)

I realize I forgot to write down that 30-2x is squared.

So it would be 2(30-2x)(-2)(x)+(30-2x)^​2

 

[JeffM]

I realize I forgot to write down that 30-2x is squared.

Yes, that was your major error. Volume of a cube-like volume should involve a cubic, not a square, right?

So it would be 2(30-2x)(-2)-4x

No.

In my first post, I worked out the formula for the volume for you in a way that leads to the simplest calculation of the derivative, but you CAN use the multiplication and chain rules to calculate the derivative.

\(\displaystyle V(x) = x(30 - 2x)^2 \implies \)

\(\displaystyle V'(x) = (1)(30 - 2x)^2 + x(2)(30 - 2x)(-\ 2) = WHAT?\)