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
Originally Posted by camherokid
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