Find dimensions to minimize cost of box

Question

A box is constructed out of two different types of metal. The metal for the top and bottom, which are both square, costs $5 per square foot and the metal for the sides costs $2 per square foot. Find the dimensions that minimize cost if the box has a volume of 50 cubic feet.

Length of base x=

Height of side z=

Answer 1

This problem is a typical optimization problem.    There is a cost function , C(x,z)

    and a constraint condition:   50 =  x² z 

 

 C(x,z) = 2 x² 5  +  4 xz 2

 

     The 2 is there because there are two surfaces costing $5/ ft²  and the 4 is there because there are four sides costing $2 /ft²  .  

 

  The way ahead is to solve the constraint condition for z     z = 50/x² 

    substitute for z in C,   to get     C =  10 x² + 400 / x

   Now C depends only on x, so the derivative of C with respect to x can be calculated and set equal to zero.

  The resulting equation can be solved for x .   The result is x =   (20)^1/3 

 

    Plugging this back into the constraint condition gives    z = 50 /(20)^2/3

 

Answer 2

Volume of box = base times height = x²z = 50

Cost = C= $5(2x²) + $2*(4xz)

solving the first eqn for z and substitute z = 50/x² into the cost equation

C = 10x² + 400/x

Find the derivative of C and set equal to zero to find minimum

C' = 20x - 400x^-2 = 0

x³ = 20

x = 2.71ft

z = 6.79ft

cost of top = $36.84

cost of each side = $36.84

 

 

Answer 3

Volume = x²y = 50

 

Therefore y = 50x^-2

 

Cost of top and bottom (2)x²($5.00), cost of the 4 sides is 4xy($2.00)   Simplify and substitute for y and we get:

Total Cost = 10x² + 8(x)(50x^-2) = 10x² + 400x^-1

 

Differentiate, set = 0 and solve

 

20x - 400x^-2 = 0

 

20x/1 = 400/x²  Cross multiply

 

20x³ = 400

 

x³ = 20

 

x = Cuberoot of 20