calculus
posted by Anonymous .
An open box of maximum volume is to be made from a square piece of cardboard, 24 inches on each side, by cutting equal squares from the corners and turning up the sides to make the box.
(a) Express the volume V of the box as a function of x, where x is edge length of the square cutouts.
(b) What are the dimensions of the box that enclose the largest possible volume? State your answer in the form length by width by height.
(c) What is the maximum volume?

let each side of the equal squares be x inches
length of box = 242x
width of box = 242x
height of box = x
a) Volume = x(242x)(242x)
b) expand the volume equation, then take the first derivative.
Set that derivative equal to zero. You will have a quadratic equation. Take the positive answer which lies between 0 and 12
c) put the answer from b) into the volume equation and evaluate. 
20>o
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