dimensions of the box to minimize cost

Question

A box with an open top has a rectangular base and has a total volume of 600m³. The length of the box is three times the width. The sides of the box are made of aluminum which costs $5/m² and the bottom is made of tin which costs $3/m². Determine the dimensions of the box that will minimize the cost of the materials

Answer 1

3.24 m by 7.85 m by 23.55 m are the dimensions that minimize cost

V=600 m^3

L=3W

V=HWL = HW(3W) = H3W^2=600

H= 600/3W^2= 200/W^2

HW = (200/W^2)(W) = 200/W

HL = L200/W^2 = 600W/W^2 = 600/W

4 sides' area = 2(HL + HW) = 2(600/W+200/W) = 1600/W

bottom area = LW = 3WW = 3W^2

cost of sides = 5(1600/W)= 8000/W

cost of bottom = 3(3W^2) = 9W^2

total cost = C = 8000/W + 9W^2

take the derivative and set it equal to zero, then solve for W

C' = -8000/W^2+18W = 0

multiply by W^2

-8000 +18W^3 = 0

W^3 = 8000/18

W = 10x cube root of 4/9 = 7.85

L=3W = 23.55

H = 600/LW = 600/184.9 = 3.24

H=3.24, L= 23.55, W= 7.85

V = HLW = 3.24x23.55x 7.85 = 599

C=5(2)(HW+HL) + 3LW = 10(25.43+ 76.30) + 3(184.9)

= 554.6 + 1017.3 = $1,571.9 = minimum cost with dimensions 3.24 by 23.55 by 7.85 meters

Answer 2

Because we are given that length = 3w and the volume constraint, we will be able to write a function first for the surface area then for the cost, solely in terms of the variable w. We can then find the derivative of the cost function, set it = 0, and solve for the ideal width to minimize the cost:

SA = lw + 2wh + 2lh and since l = 3w, SA = 3w² + 2wh + 6wh = 3w² + 8wh ,

V = lwh = 3w²h = 600 and h = 200/w² plugging that in above we get SA = 3w² +1200w^-1

Cost: C(w) = 9w² + 6,000w^-1

C^'(w) = 18w - 6,000/w² = 0. w³ = 1000/3 and w = 10 / ³√3 l = 10·³√9 and h = 2·³√9