Question

Suppose we want to build a rectangular storage container with open top whose volume is $$12 cubic meters. Assume that the cost of materials for the base is$$‍12 dollars per square meter, and the cost of materials for the sides is $$8 dollars per square meter. The height of the box is three times the width of the base. What’s the least amount of money we can spend to build such a container?

Answer 1

Answer:

w = w    L = 2w    h = h

Volume:     V = Lwh

                 10 = (2w)(w)(h)

                 10 = 2hw^2

                   h = 5/w^2

Cost:    C(w) = 10(Lw) + 2[6(hw)] + 2[6(hL)])

                      = 10(2w^2) + 2(6(hw)) + 2(6(h)(2w)

                      = 20w^2 + 2[6w(5/w^2)] + 2[12w(5/w^2)]

                      = 20w^2 + 60/w + 120/w

                      = 20 w^2 + 180w^(-1)

            C'(w) = 40w - 180w^(-2)

Critical numbers:

          (40w^3 - 180)/w^2 = 0

                      40w^3 -180 = 0

                              40w^3 = 180

                                  w^3 = 9/2

                                      w = 1.65 m

                                       L = 3.30 m

                                       h = 1.84 m

Cost:  C = 10(Lw) + 2[6(hw)] + 2[6(hL)])

              = 10(3.30)(1.65) + 2[6(1.84)(1.65)] + 2[6(1.84)(3.30)])

              = $165.75                       cheapest cost

Step-by-step explanation: