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 $

sas (side angle side)

aa(angle angle)

sss(side side side)

sas you have to find two sides with an angle in between

aa you find two angles because it will always add up to 180

sss find all sides

step-by-step explanation:

the answer would be 0.54