CALCULUS HW HELP PLEASE

Question

A box with an open top is to be constructed from a 13 ft by 12 ft rectangular piece of cardboard by cutting out squares or rectangles from each of the four corners, as shown in the figure, and bending up the sides. One of the longer sides of the box is to have a double layer of cardboard, which is obtained by folding the side twice. Find the largest volume (in ft³) that such a box can have. (Round your answer to two decimal places.)

Answer 1

The volume is going to be width • length • height. Since we are doubling the long side, we will cut two squares out of the top and bottom of one side, letting each side be x. The other side, we will cut rectangles with one side equal to 2x along the width and x along the length. The height, after folding will be x.

So, Volume, V = (12 - 3x)(13-2x)x = 156x - 63x²+ 6x³.

To maximize this cubic equation, take the derivative of V.

V ' = 156 -126x + 18x².

Set this to zero and solve for x. x = 1.6, 5.39 using quadratic equation

Turns out 1.6 gives a maximum and 5.39 is a minimum. Pick 1.6

V = (12-4.8)(13-3.2)(1.6) = (7.2)(9.8)(1.6) = 112.9 ft³

Answer 2

13*12

2(13-3*1)(12-2*1)=200cu/ft