Arectangular storage container with a lid is to have a volume of 14 m3. the length of its base is twice the width. material for the base costs $7 per m2. material for the sides and lid costs $14 per m2. find the dimensions of the container which will minimize cost and the minimum cost.

C = 42 7^(2/3) + 84 * 7/7^(1/3) = 126 7^(2/3)

Step-by-step explanation:

Let x be the length of the base, y its width and z the height.  Then x=2y,

Volume = V = xyz = 2y^2 z = 14 so z = 7/y^2

Then the cost is C = 7 xy + 14 (xy + 2xz + 2yz) = 21 xy + 28 xz + 28 yz =

= 42 y^2 + 56 y z + 28 yz = 42 y^2 + 84 yz = 42 y^2 + 84 y (7/y^2) =

= 42 y^2 + 84*7/y and so

dC/dy = 84 y - 84*7/y^2 = 0 and so y^3 = 7 and y = 7^(1/3) and therefore

x = 2y = 2 7^(1/3) and z = 7 / 7^(2/3) = 7^(1/3).  The minimum cost is then

C = 42 7^(2/3) + 84 * 7/7^(1/3) = 126 7^(2/3)