Optimization Problems

Question

A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width. Material for the base costs $20 per square meter. Material for the sides costs $12 per square meter. Find the cost of materials for the cheapest such container. (Round your answer to the nearest cent.)

Answer 1

w: width of base

h: height of box

V = 2w·w·h = 2hw² = 10 ; h = 5 / w²

Surface Area_base = 2w² ; Surface Area_sides = 2hw + 4hw = 6hw

Cost = 40w² + 72hw

C(w) = 40w² + 360w^-1 ; w > 0

C^'(w) = 80w - 360w^-2 = 0

2w³ - 9 = 0

w = ³√(9/2)

Min cost: C(³√(9/2)) = ...

Answer 2

We know that lwh = 10

Also, l = 2w

Therefore, (2w)wh = 10

2w²h = 10

w²h = 5

This is our constraint

Now we have to minimize the cost

Cost = 20l*w + 12(2)w*h + 12(2)l *h

Let's replace the ls with 2w

Cost = 20w*w + 24w*h + 24(2w) * h

Cost = 20w² + 24wh + 48wh

Cost = 20w² + 72wh

Now we have to minimize the cost. Solve for a variable in the constraint.

h = 5/w²

Now we plug that into the cost function

Cost = 20w² + 72w(5/w²)

Cost = 20w² + 144/w

Take the derivative and set it equal to 0

Cost' = 40w -144/w² = 0

8/w²(5w³ - 18) = 0

5w³ - 18 = 0

5w³ = 18

w³ = 3.6

w = ³√3.6

Therefore, l = 2³√3.6

h = 5/³√3.6²

Lastly, use the formula for cost to find the cost