Find the length and width of such a container for which the cost of materials is minimum.

Question

A rectangular storage container without a lid is to have a volume of 10 meter^3. The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the length and width of such a container for which the cost of materials is minimum.

Answer 1

L=2W for the base

area of the base = LW = 2WW = 2W^2

at a cost of 10(2W^2) = 20W^2

sides cost 6 dollars,

4 sides, 2 HW, 2 HL

2HW+2HL = 2HW + 2H(2W) = 6HW

cost = 6(6HW) = 36HW

Volume = HWL = 10 m^3

H=10/WL = 10/W2W = 5/W^2

C=Cost = 36HW +20W^2 = 36(5/W^2)W + 20W^2 = 180/W + 20W^2

C' = 40W -180/W^2 = 0

40W^3 =180

W^3 = 180/40= 18/4=9/2

W = cube root of 9/2 = about 1.651 meters base width

L = 2W = 3.302 meters base length

H = 5/W^2 = 5/1.651^2 = 1.834 meters high

Answer 2

Let x is width, then 2x is length and h is height. Then cost is C(x) = 10·2x² + 6·6xh = 20x² + 36xh^.

Volume v = 2x²h = 10; h = 5/x². C(x) = 20x² + 180/x;

C'(x) = 40x - 180/x² = 0; 40x³ = 180; x = 4.5^1/3 = 1.651 m; 2x = 3.302 m

and this is min because C''(x) = 40 + 360/x³, C''(4.5^1/3) = 40 + 360/4.5 = 120 > 0;