1. Let side of square base of the box = x cm
and height of the box = y cm.
Volume = x²y
4000 = x²y
y = 4000/x² ....(1)
The material used for the box Surface Area = Areaof base + 4 times area of sides
Surface Area, A= x² + 4xy
Plug in value from (1) to get
A = x² + 4x(4000/x²)
A = x² + 16000/x
To find optimal value, find derivative and equate to 0
A' = 2x - 16000/x²
0 = 2x - 16000/x²
16000/x² = 2x
x³ = 8000
x = 20 cm
y = 4000/(20)² = 10
Dimension of the box is 20 cm x 20 cm x 10 cm
2. Let width of the base of the box = x meter
then length of the base = 2x meter
and height of the box = y meter
Volume of the box = x(2x)y = 2x²y
10 = 2x²y
y = 5/x² ....(1)
Cost of box material, C = cost of base + cost of sides
C = 20(2x²) + 12(2xy+4xy)
C = 40x² + 72xy
C = 40x² + 72x(5/x²) .........from (1)
C = 40x² + 360/x
C' = 80x - 360/x²
0 = 80x - 360/x²
360/x² = 80x
x³ = 4.5
x = 1.65 m
y = 5/(1.65)² = 1.84 m
Cost, C = 40x² + 360/x = 40(1.65)² + 360/1.65 = $327.08
