1. A box with a square base and open top must have a volume of 4,000 cm3. Find the dimensions of the box that minimize the amount of material used.

2. A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width. Material for the base costs $20 per square meter. Material for the sides costs $12 per square meter. Find the cost of materials for the cheapest such container.

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