A box with a square base and open top must have a volume of 171500 cm^3

Question

A box with a square base and open top must have a volume of 171500 cm^3. We wish to find the dimensions of the box that minimize the amount of material used.

First, find a formula for the surface area of the box in terms of only x, the length of one side of the square base.

First, find a formula for the surface area of the box in terms of only xx, the length of one side of the square base.

A(x)=

Next, find the derivative, A'(x)

A'(x)=

The critical value is x=

Answer 1

A box with a square base and open top must have a volume of 171500 cm^3. We wish to find the dimensions of the box that minimize the amount of material used.

Detailed Solution:

Volume of Box = v = Area of Base * Height x = dimensions of base, y = dimensions of height

v = (x)(x) * (y) = x²y = 171500 ==> y = 171500/x²

Surface Area = s = Area of Base + 4*(Area of Sides) = (x)(x) + 4*(xy) = x² + 4xy

s = x² + 4xy <== Minimize this Surface Area

s = x² + 4x(171500/x²) <== Substitute for y

s = x² + 4(171500)/x = x² + 4(171500)x^-1

ds/dx = 2x − 4(171500)x^-2 = 0 <== take 1st derivative and set it to "0"

2x − 4(171500)/x² = 0

(2x − 4(171500)/x² = 0) (x²) <== Multiply both sides by x²

2x³ − 4(171500) = 0

2x³ = 4(171500)

x³ = 2(171500)

(x³)^(1/3) = (2(171500))^(1/3)

x = 70 cm

y = 171500/x² = 171500/(70)² = 35

y = 35 cm

The dimensions of the box that minimize the amount of material used are: (x, y) = (70 cm, 35 cm)

The minimum amount of material used for the box:

s = x² + 4xy = (70)² + 4(70)(35) = 14700 cm² <== minimum amount of material used for the box

To conform it is minimum/minimization, take second derivative

s'' = 2 + (2)(4)(171500)x^-3 <== 2nd derivative

s'' = 2 + (2)(4)(171500)/x³ <== 2nd derivative is always positive (>0) for positive x; thus Concave up ==> Minimum. That is: (x, y) = (70 cm, 35 cm) is minimum - the dimensions of the box that minimize the amount of material used.The minimum amount of material used for the box: s = 14700 cm²