"a rectangular storage container with a lid is to have a volume of 12 m^3. the length of its base is twice the width. material for the base costs $6 per m2. material for the sides and lid costs $12 per m2. find the dimensions of the container which will minimize cost and the minimum cost."

The dimensions of the box are 3.634m of length, 1.817m of width, and 1.817m of height with a cost of $356.61

Step-by-step explanation:

The volume of the box is V = L·W·H

Since L=2W then V=2W²H=12m³

The area of the base is B = L·W = 2W², so the cost of the base is

Cb=6·2W²=12W²

The area of one long side is S₁=L·H and one wide side is S₂=W·H so the area of the sides is

S=2L·H+2W·H = 4W·H+2W·H = 6W·H

so the cost of the sides is

Cs=12·6W·H=72W·H

The area of the lid is A=L·W, so the cost of the lid is

Cl=12·L·W = 12·2W·W = 24W²

The total cost is then

C= 12W²+72W·H+24W² = 36W²+72W·H

Isolating H from the volume we get

So the cost becomes:

finding the first derivative:

We find the roots (C'=0) to find maximum and minimum:

evaluating the second derivative of the cost:

evaluating the root on C'' we find:

which is a positive number, so the root W=∛6=1.817m is a minimum for the function of cost.

Being W=∛6 :

and

Which are the dimensions of the box for a minimum cost of:

$15.40 per mile

step-by-step explanation:

you have to round up to 4 because 6 is over 5 so you have to round the 3 up to 4