How do i solve it

Question

Here is the question: A closed rectangular box with a volume of 16ft^3 is made from two kinds of materials. The top and bottom are made of material costing 10 cents per square foot and the sides from material costing 5 cents per square foot. Find the dimensions of the box so that the cost of materials is minimized.

Answer 1

V=16 = LWH where L=length, W=width, and H = height of the rectangular box

H=16/LW

Area of top and bottom is 2LW

Area of the sides is the sum of 4 side areas = 2LH + 2WH

cost of top & bottom is 10(2LW) = 20LW

cost of 4 sides = 5(2)(L+W)H= (10L + 10W)16/LW==160(L+W)/LW= 160/W + 160/L

total cost = C = 20LW + 160/W + 160/L

Let L=W

C = 20L^2 + 320/L

C'= 40L -320/L^2 = 0

L -8/L^2 = 0

L^3 = 8

L = 2 = W

16 =LWH

H = 16/4 = 4

minimum cost = 10(2)(L^2) + 320/L = 80 + 160 = $240

dimensions of the box are 2 x 2 x 4 feet

IF the material had cost the same for top, bottom and sides, the dimensions would have be H=W=L. It would have been a cube, with each dimension = cube root of 16

But because the sides are cheaper than the top&bottom, half as cheap, the height is twice as big as the length or width.

the square top and bottom minimize cost for the top and bottom, so L=W