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
 

true !

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congruent segments

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