SOLUTION: 1. A rectangular piece of cardboard can be turned into an open box by cutting away identical squares from each corner and folding up the resulting flaps. The cardboard is 12 inche

Question 1143072: 1. A rectangular piece of cardboard can be turned into an open box by cutting away identical squares from each corner and folding up the resulting flaps. The cardboard is 12 inches long and 12 inches wide.
a) Write an expression for the length of the box in terms of x (the length of each square you cut away).
b) Write an expression for the width of the box in terms of x.
c) Write an expression for the Volume of the box in terms of x.
d) Find the dimensions of the box that will yield the maximum volume.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
, the volume of the box formed
height is x and length and width each is 12-2x.
Can you find solution to part (d) yourself?
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.

The dimensions of the box are (12-2x) inches (each the length and the width of the base). The height of the box is x inches. The volume of the box is V(x) = x*(12-2x)*(12-2x) = x*(144 - 48x + 4x^2) = 4x^3 - 48x^2 + 144x. To determine the maximum of the function V(x), take the derivative V'(x) = 12x^2 - 96x + 144 and equate it to zero 12x^2 - 96x + 144 = 0. It gives x^2 -8x + 12 = 0 Factor left side (x-6)*(x-2) = 0. The two roots are x= 6 and x= 2. The volume at x= 6 is equal to zero, so this root, although provides the local minimum, does not give a real solution. The volume at x= 2 is V(2) = 2*(12-2*2)*(12-2*2) = 2*8*8 = 128 cubic inches. It is the real maximum and the real solution to the problem. Plot V(x) = 4x^3 -48x^2 + 144x.

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