Question

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?

Answer 1

bear with me.. solving

Answer 2

Surface area is area of the bottom: x^2 + 4 sides: 4*xy
Surface area = 48ft^2 = x^2 + 4xy
Volume is x^2*y:
V = x^2 * y

Answer 3

Solve for y in the surface area:
48 = x^2 + 4xy
48-x^2 = 4xy
(48-x^2)/4x = y

Answer 4

y = (48 - x^2) / 4x
V = x^2 * y = x^2 * (48 - x^2) / 4x = x ( 48-x^2) / 4

Answer 5

Simplify a bit:
V = 48x/4 - x^3/4 = 12x - x^3/4
Take the derivative and set it equal to 0:
V ' = 12 - 3x^2 / 4 = 0

Answer 6

Solve the equation:
12 - 3x^2 / 4 = 0
-3x^2/4 = -12
x^2/4 = 4
x^2 = 4*4 = 16

Answer 7

x = + or - 4, but in this case it's a plus because the edge of the box cannot be negative.. anyway let's double check if it's min or max. To do so, take the 2nd derivative:
V" = -6x / 4 = - 3x / 2. Which is positive if x = -4 and negative if x = 4. Maximum occurs when 2nd derivative is negative therefore x = 4 is max

Answer 8

Okay so now that you have x = 4, find y:
y = (48 - x^2) / 4x
y = (48 - 16) / 16 = 32/16 = 2
So dimensions are:
The sides of the base must be equal to 4, and the height of the box must be 2. In which case the volume = area of the base * height:
V = 4^2 * 2 = 16 * 2 = 32

Answer 9

Hope this helped =)

Answer 10

by the way units are:
x = 4ft
y = 2ft
V = 32ft^3