# Thread: Optimization Problem3

1. ## Optimization Problem3

Rectangle storage container with an open top is to have a volume 10 m^3.
The length of its base is twice the width.
Material for the base costs $10 /m^2. Material for the sides costs$6 /m^2.
Find the cost of materials for the cheapest such container.

2. Originally Posted by camherokid
Rectangle storage container with an open top is to have a volume 10 m^3.
The length of its base is twice the width.
Material for the base costs $10 /m^2. Material for the sides costs$6 /m^2.
Find the cost of materials for the cheapest such container.
let the width of the box be x, then the length is 2x
let the height of the box be y

we want a box of the smallest surface area to minimize the cost.

we have V = lwh
=> V = x(2x)(y) = 2x^2 * y
we are told V = 10
=> 2x^2 * y = 10
=> y = 10/2x^2 = 5/x^2

now the surface area is given by
S = area of base + area of the sides
=> S = 2x^2 + 6xy
=> S = 2x^2 + 6x(5/x^2)
=> S = 2x^2 + 30x^-1

=> S' = 4x -30x^-2
for min surface area, set S' = 0
=> 4x - 30x^-2 = 0
=> 4x^3 - 30 = 0 .............multiplied through by x^2
=> 4x^3 = 30
=> x^3 = 30/4 = 15/2
=> x = cuberoot(15/2) ~= 1.9574

but y = 5/x^2 = 5/(1.9574)^2 ~= 1.305

so for min surface area: x = 1.9574, y = 1.305

so the area of the base = 2x^2 = 7.6628
so the cost of the base is $76.63 so the area of the sides is 6xy = 6(1.9574)(1.305) = 15.3264 so the cost of the sides is$91.96

so the cost of the material needed to make such a box is $91.96 +$76.63 = $168.59 ~=$169