A rectangular box is to have a square base and a volume of 40 ft^3. If the material for the base costs $0.29 per square foot, the material for the sides costs $0.05 per square foot, and the material for the top costs $0.21 per square foot, determine the dimensions of the box that can be constructed at minimum cost. length ft width ft height ft Can anyone help me?
Solution
let the dimensions of base = x ft
let the height be h cm
volume, V = x^2 h = 20
h = 20 / x^2
area of base = x^2
area of top = x^2
area of 4 sides = 4xh
Total Cost, C = 0.3x^2 + 0.2 x^2 + 0.1(4xh)
C = 0.5x^2 + 0.4 xh
substitute h = 20 /x^2
C = 0.5 x^2 + 8 / x
differentiating
dC/dx = x - 8 x^2
inorder to minimize cost , equate dC/dx to 0
x - 8/x^2 = 0
x^3 = 8
x = 2 ft
h = 20 /4 = 5
so dimensions of box are 2 x 2 x 5 ft
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A rectangular box is to have a square base and a volume of 40 ft^3- If.docx
1. A rectangular box is to have a square base and a volume of 40 ft^3. If the material for the base
costs $0.29 per square foot, the material for the sides costs $0.05 per square foot, and the
material for the top costs $0.21 per square foot, determine the dimensions of the box that can be
constructed at minimum cost. length ft width ft height ft Can anyone help me?
Solution
let the dimensions of base = x ft
let the height be h cm
volume, V = x^2 h = 20
h = 20 / x^2
area of base = x^2
area of top = x^2
area of 4 sides = 4xh
Total Cost, C = 0.3x^2 + 0.2 x^2 + 0.1(4xh)
C = 0.5x^2 + 0.4 xh
substitute h = 20 /x^2
C = 0.5 x^2 + 8 / x
differentiating
dC/dx = x - 8 x^2
inorder to minimize cost , equate dC/dx to 0
x - 8/x^2 = 0
2. x^3 = 8
x = 2 ft
h = 20 /4 = 5
so dimensions of box are 2 x 2 x 5 ft
