Caluclus, applications of the derivative problem?

Question

An open box is to be made from a square piece of cardboard whose sides are 8.00 in. long, by cutting equal squares from the corners and bending up the sides. Determine the side of the square that is to be cut out so that the volume of the box may be a maximum.

Answer 1

Let x = the length of the side of the square that is to be cut from each corner of the square piece of cardboard (which also will be the height of the box). Then the sides of the cardboard now measure 8-2x . The volume of the rectangular box as a function of x then is V(x) = x(8-2x)^2 which is the function that needs to be maximized.

V(x) = 4x^3-32x^2+64x for values of x>0

The first derivative of V(x) = 12x^2-64x+64. This needs to be set equal to zero so 12x^2-64x+64 = 4(3x^2-16x+16) = 4(3x-4)(x-4) = 0. Therefore x = 4/3 or x = 4.
One of these values will produce the maximum for V(x). To determine which will do this we need to detemine the behavior of the first derivative on either side of these values (or the 2nd derivative test could also be used). To the left of 4/3 the derivative is positive, to the right of 4/3 the derivative is negative, which implies that x = 4/3 is a maximum for the volume function. Checking the other value in a similar manner we see that to the left of 4 the derivative is negative and to the right of 4 it is positive which means that x = 4 will be a minimum.

The side of the square that needs to be cut out then is 4/3 in, which will make the volume of the box 1024/27 or 37.93 cubic inches.

Answer 2

you need 5 equal square pieces to make an open square box.
To cut square pieces from a square cardboard 8" x 8", you can either make up 4 squares of 4' x 4" (not enough) or 9 squares of 8/3" x8/3", so the side to get max volume is 8/3 = 2&2/3 "

Answer 3

Let squares be of side x inches.
Base of box is of side = (8 - 2x) inches.
height of box is x inches
V(x) = (8 - 2x)² ( x )
V(x) = (64 - 32x + 4x²) (x)
V(x) = 64x - 32x² + 4x³
V `(x) = 64 - 64x + 12x² = 0 for max.
V `(x) = 3x² - 16x + 16 = 0
(3x - 4) (x - 4) = 0
x = 4 / 3 , x = 4
x = 4 / 3 is valid value
4/3 inches is side of square to be cut out.