maximum volume, help?

Question

Find the maximum volume of a box that can be made by cutting out squares from the corners of an 8-inch by 15-inch rectangular sheet of cardboard and folding up the sides.

Answer 1

You don't indicate whether this box has a top.

Suppose it does not. If the height of the box is h, then its volume will be

V = (15-2h)*(8-2h)*h = 4h³ - 46h² + 120h

dV/dh = 12h² - 92h + 120; for a maximum,

12h² - 92h + 120 = 0,
3h² - 23h + 30 = 0
(3h - 5 )(h - 6) = 0

h = 5/3, h = 6

d²V/dh² = 24h - 92,

which is negative if 92 > 24h; h < 3.83.

V(5/3) = 4(125)/27 - 46(25)/9 + 120(5)/3
V(5/3) = 18.52 - 127.78 + 200 = 90.74

The box is 11-2/3 by 4-2/3 by 5/3 with a volume of 90.74 in³

Answer 2

Let the length of the sides of the cut-out square be x.

The length would be 8-2x, the width would be 15-2x and the height would be x.

V = (8-2x)(15-2x)(x)
V = 120x-46x^2+4x^3
Take the derivative of V with respect to x, and set it to 0.
V' = 120-92x+12x^2 = 0
Using the quadratic formula or factoring or whatever have you, you get x = 5/3 and 6. However, the answer is 5/3 because x=6 actually gives the minimum and can be verified with second derivative. (and cutting 6 inches squares is impossible in this problem)

So V = (8-10/3)(15-10/3)(5/3) = 2450/27 inches cubed

Answer 3

The volume will be x(8-2x)(15-2x)
So V = 4x^3 -46x^2 +120x
dV/dx = 12x^2 -92x +120
x = 6 or x = 1 2/3
But x cannot = 6 because then the 8 inch side would be too small. Thus x = 1 2/3 inches.
V= 4(5/3)^3 -46(5/3)^2 +120(5/3) = =90.74 in^3