Calculus [optimization] question?

Question

The question is:

Consider the following problem: A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have. Let x denote the length of the side of the square being cut out. Let y denote the length of the base.

I have the piece of cardboard all drawn out, along with the 4 squares, whose lengths are X.

Where do I go from here? Is the Y the base of the ENTIRE piece of cardboard or the bases of the 4 squares?

Answer 1

ok,

y is the length of the bottom of the box when you fold up the sides

x is the height of the box when you fold up the sides

y = 3 - 2x (relationship between y and x)

V = x * y^2 (volume of the box)

V = x (3-2x)^2 (substituting the relationship above for y)

V = x (9-12x+4x^2) (expanding the square)

V = 9x - 12x^2 + 4x^3 (performing distribution)

take first derivative

V' = 9 -24x + 12x^2

set equal to zero

0 = 9 - 24x + 12x^2

divide by 3 to remove the greatest common factor

0 = 3 - 8x + 4x^2

factor the equation

0 = (2x -3)(2x-1)

use zero factor property

2x-3 = 0

2x = 3

x = 3/2

or

(2x-1) = 0

2x = 1

x =1/2

one of these is smallest box, other is largest box

V = 9x - 12x^2 + 4x^3

V = 9(3/2) - 12(3/2)^2 + 4(3/2)^3 = 0

V =9(1/2) - 12(1/2)^2 + 4(1/2)^3 = 2

largest possible box has a volume of 2 ft cubed <----ANSWER

incidentally the height is 1/2 ft and the length is 2ft