can't find an equation to use for this max/min problem?

Question

can't find an equation to use for this max/min problem

an open box is to be made from a sheet of card board with area 24 ft. squared by cutting out even squares (sides x ft.)from the corners and folding up. what are the dimensions of the box for the maximum volume?

Answer 1

A = LxW = 24
A for box = (L-2x)*(W-2x)
V of Box = x (L-2x) *(w-2x)

V = x * (L*W -2XW + 4x^2 -2xL)
V = LWx -2wx^2 + 4x^3 -2x^2L

Take derivative of v in terms of x and equate to zero to get the maximum
dV/dx = 0 to get maximum volume;

0 = LW - 4Wx + 12X^2 -4xL

0 = 24 -4x(W+L) + 12X^2

0 = 6 -x*(W+L) + 3X^2
you need at least L or W to finish this problem :)

Answer 2

The height will be x.
The width will be the width (24) minus 2x.
The length will also be 24 - 2x.

So the formula for the volume will be:
f(x) = x (24 - 2x)(24 - 2x)

Or multiply it out:
f(x) = 4x^3 - 96x^2 + 576x

Edit: For others that are viewing this as 24 sq. ft, I disagree with your reasoning. It's obvious that with this logic you introduce another unknown... I don't think that was what was intended.

Indeed if you check other references you find that there is a difference between 24 ft squared (24' x 24') vs. 24 sq. ft (e.g. something like 4' x 6' or 3' x 8').

I stick by my original answer.

Continuing with the problem:
f(x) = 4x^3 - 96x^2 + 576x

To figure the maximum, take the first derivative:
f'(x) = 12x^2 - 192x + 576

Now you can set this to zero:
0 = 12x^2 - 192x + 576

Divide both sides by 12:
0 = x^2 - 16x + 48
0 = (x - 12)(x - 4)

So either x = 12 or x = 4
x = 12 is the minimum (volume of zero)
x = 4 is the maximum (volume of 4 x 16 x 16 = 1,024 cubic feet*)

*not to be confused with 1,024 feet cubed. And personally I think math books should avoid confusion and never say something like 24 ft. squared because it can easily lead someone astray. If they had just said a square piece of cardboard 24' on side, no one would have been confused. anf3rnee23, when you get your test back, please tell your teacher that we all agree that the wording is confusing and it shouldn't count against you if you got it wrong on the test. You obviously understand the way to solve it, if you just were able to understand the question.

Answer 3

Were you given any other information? For example, is the sheet of cardboard perhaps a SQUARE-shaped sheet? Otherwise you end up with two variables.

Let L be the original length of the sheet before the the cut. Then the width is 24/L, since the original area (length times width) must be 24 square feet. When the four x-by-x squares are taken from each corner and the flaps are folded upwards, this makes the box have a depth of x, a length of L - 2x, and a width of (24/L) - 2x. It might be helpful to make a diagram to see why.

This means the volume of the box is
x(L - 2x)(24/L - 2x) =
x(24 - 48x/L - 2xL + 4x^2) =
x(24 - (48/L + 2L)x + 4x^2) =
24x - (48/L + 2L)x^2 + 4x^3

ASSUME that L is actually a constant. Then take the derivative with respect to x, and set it equal to zero:
24 - 2(48/L + 2L)x + 12x^2 = 0
6x^2 - (48/L + 2L)x + 12 = 0
3x^2 - (24/L + L)x + 6 = 0

Use the quadratic formula to solve this for x in terms of L. Again, you end up with two variables here. You can then use these values to find the dimensions of the box: x, (L - 2x), and (24/L - 2x).

This makes some sense if you think about it: the squares you'd cut out of a 6x4 piece of cardboard might not be the same that you'd cut out of an 8x3 piece of cardboard.