A rectangular storage container with an open top is to have a volume of 14 cubic meters. The length of its base is twice the width. Material for the base costs 10 dollars per square meter. Material for the sides costs 8 dollars per square meter. Find the cost of materials for the cheapest such container.

C(min)  =  277.95 $

Container dimensions:

x = 2.822 m

y = 1.411 m

h = 3.52 m

Step-by-step explanation:

Let´s call x  and  y the sides of the rectangular base.

The surface area for  a rectangular container is:

S = Area of the base (A₁) +  2 * area of a lateral side x (A₂) + 2 * area lateral y (A₃)

Area of the base is :

A₁ =  x*y       we assume, according to problem statement that

x  =  2*y      y  = x/2

A₁  =  x²/2

Area lateral on side x

A₂  = x*h           ( h  is the height of the box )

Area lateral on side y

A₃ = y*h        ( h  is the height of the box )

s = x²/2  + 2*x*h + 2*y*h

Cost  =  Cost of the base + cost of area lateral on x + cost of area lateral on y

C = 10*x²/2  + 8* 2*x*h  + 8*2*y*h

C as function of x is:

The volume of the box is:

V(b) = 14 m³  = (x²/2)*h       28 = x²h      h = 28/x²

C(x) = 10*x²/2  + 16*x*28/x² + 16*(x/2)*28/x²

C(x)  =  5*x²  +  448/x  + 224/x

Taking derivatives on both sides of the equation we get:

C´(x)  =  10*x  -  448/x² - 224/x²

C´(x)  =  0             10x  -  448/x² - 224/x² = 0     ( 10*x³ - 448 - 224 )/x² = 0

10*x³ - 448 - 224 = 0       10*x³ = 224

x³  =22.4

x = ∛ 22.4

x = 2.822 m

y = x/2  =  1.411 m

h = 28/x²  =  28 /7.96

h =  3.52 m

To find out if the container of such dimension is the cheapest container we look to the second derivative of C

C´´(x) = 10 + 224*2*x/x⁴

C´´(x)  =  10 + 448/x³    is positive then C has a minimum for x = 2.82

And the cost of the container is:

C = 10*(x²/2) +  16*x*h + 16*y*h

C = 39.82 +  158.75  + 79.38

C =  277.95 $

2 5/8 is the final answer! : )

rounding to the nearest thousandth would be 452,000