A has a square base of side x and height y

Question

A box has a square base of side x and height y.

(a) Find the dimensions x, y for which the volume is 12 and the surface area is as small as possible.

(b) Find the dimensions for which the surface area is 20 and the volume is as large as possible.

Answer 1

yx^2 = 12 = volume= V

y = 12/x^2 = height

surface area = 2x^2 + 4xy = SA = 2x^2 + 4x(12/x^2) = 2x^2 +48/x if the top of the box is closed

take the derivative, set = zero, then solve for x

SA' = 4x -48/x^2 = 0

x = 12/x^2

x^3 = 12

x= 12^1/3 = about 2.3

y= about 12/2.3^2 = about 2.3

SA = 2x^2 +4xy = 6(2.3)^2 = about 31.8 = minimum surface are if volume = 12

x=y = about 2.3

minimum surface area for a given volume is a cube

IF SA = 20 = 2x^2 + 4xy

4xy = 20-2x^2

2xy = 10-x^2

y = (10-x^2)/2x

V = yx^2= [(10-x^2)/2x][x^2] = (10x-x^3)/2= 5x -x^3/2

take the derivative and set = 0, then solve for x

V'= 5 -(3/2)x^2 = 0

x^2 = 5(2/3) = 10/3

x = sqr(10/3) = about 1.8

y= about (10-1.8^2)/2(1.8) = 5/1.8 -0.9 = 2.8-.9= 1.8

largest volume is a cube = about 1.8^3 = about 5.8

max V= about 1.8^3 = 5.8