Arectangular storage container with an open top is to have avolume
20m3 . the length of its base is twice the width. the
material for the base costs $5 per square meter. the materialfor
the sides costs $9 per square meter. find the minimum cost tobuild
such container.

Dimensions of the container:

x = 3 m

y = 6 m

h = 1.1 m

C(min) = 270 $

Step-by-step explanation:

Volume of storage container

V = 20 m³

Let  "y" be the length    and  "x" the width   then y = 2*x

V = x*y*h    ⇒  V = 2*x²*h      ⇒   20 = 2*x²+h      ⇒  h  =  10/ x²

Costs:

Total cost = cost of base  ( 5*2*x² )  + cost of side with base x  ( 2*9*x*h) +

cost of side witn base y =2x   (2*9*2x*h)

C(t)  = 10*x²  +  18*x*h  +  36*x*h

C(x)  = 10x²  + 54*x*10/x²      ⇒  C(x)  10*x²   +  540 /x

Taking derivatives on both sides of the equation we get:

C´(x)  =  20*x  -  540/x²

C´(x)  =  0        ⇒  20*x  -  540/x²  =  0   ⇒  2x  - 54/x²   = 0

2x³ - 54  = 0

x³  = 27       x  =  3 m

Then  y  =  2*x    ⇒ y = 2*3   y = 6     and  h  =  10 / x²       h  = 1.1 m

And the minimum cost is

C (min)  =  10*x²  +  540/x     ⇒      C (min)  =  90 + 180

C(min) = 270 $

d? maybe

step-by-step explanation: