Calculus problems

PROBLEM: An open rectangular box with square base is to be made from 48 ft.² of material. What dimensions will result in a box with the largest possible volume ?

SOLUTION  : Let variable x be the length of one edge of the square base and variable y the height of the box.

The total surface area of the box is given to be

48 = (area of base) + 4 (area of one side) = x² + 4 (xy) ,

so that

4xy = 48 - x²

or

 .

We wish to MAXIMIZE the total VOLUME of the box

V = (length) (width) (height) = (x) (x) (y) = x² y .

However, before we differentiate the right-hand side, we will write it as a function of x only. Substitute for y getting

V = x² y

= 12x - (1/4)x³ .

Now differentiate this equation, getting

V' = 12 - (1/4)3x²

= 12 - (3/4)x²

= (3/4)(16 - x² )

= (3/4)(4 - x)(4 + x)

= 0

for

x=4 or x=-4 .

But  since variable x measures a distance and x > 0 . Since the base of the box is square and there are 48 ft.² of material, it follows that . See the adjoining sign chart for V' .

If

x=4 ft. and y=2 ft. ,

then V = 32 ft.³ is the largest possible volume of the box.