We have to find the cost of materials for the cheapest container. It has a volume of 10 m^3. V = 10 m^3; V = L W H; where L is length, W is width and H is height. We know that L = 2 W; 10 = 2 W^2 H ; H = 10 / 2W^2 ; H = 5 / W^2. Base Area = L W = 2 W W = 2 W^2. Side Area = 5 / W^2 ( 2 W + 4 W ) = 5 / W^2 6 W = 30 / W. The Cost: C ( W ) = 5 2 W^2 + 3 30 / W = 10 W^2 + 90 / W. We have to find the derivative of the cost. C(W) = 20 W - 90 / W^2 = ( 20 W^3 - 90 ) / W^2. Then: C ( W ) = 0; 20 W^3 - 90 = 0; 20 W^3 = 90; W^3 = 4.5 ; W = 1.651. Finally the cost for the cheapest container is: C ( 1.651 ) = 10 ( 1.651 )^2 + 90 / 1.651 = 10 * 2.72568 + 54.5157 = 27.2568 + 54.5157 = 81.7725 ( $81.77 to the nearest cent ). Answer: The cost of materials for the cheapest such container is $81.77.
