Answer:
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
[tex]H=\frac{6m³}{W^2}[/tex]
So the cost becomes:
[tex]C=36W^2+72W(\frac{6}{W^2})=36W^2+\frac{432}{W}[/tex]
finding the first derivative:
[tex]C'=72W-\frac{432}{W^2}[/tex]
We find the roots (C'=0) to find maximum and minimum:
[tex]0=72W-\frac{432}{W^2}[/tex]
[tex]72W=\frac{432}{W^2}[/tex]
[tex]W^3=\frac{432}{72}[/tex]
[tex]W=\sqrt[3]{\frac{432}{72}}=\sqrt[3]{6}=1.817[/tex]
evaluating the second derivative of the cost:
[tex]C''=72+\frac{864}{W^3}[/tex]
evaluating the root on C'' we find:
[tex]72+\frac{864}{(\sqrt[3]{6})^3}=72+\frac{864}{6}=216[/tex]
which is a positive number, so the root W=∛6=1.817m is a minimum for the function of cost.
Being W=∛6 :
[tex]H=\frac{6}{(\sqrt[3]{6})^2}=1.817m[/tex]
and
[tex]L=2W=2\sqrt[3]{6}=3.634m[/tex]
Which are the dimensions of the box for a minimum cost of:
[tex]C=36W^2+\frac{432}{W}=36(\sqrt[3]{6})^2+\frac{432}{\sqrt[3]{6}}=356.61[/tex]
