Answer
The minimum possible cost of materials is $~~\$163.54$
Work Step by Step
Let $x$ be the width of the base, let $2x$ be the length of the base, and let $h$ be the height of the container.
Note that $x \geq 0$ and $h \geq 0$
We can express $h$ in terms of the volume $V$ and $x$:
$V = 2x^2h$
$h = \frac{V}{2x^2}$
We can write an expression for the cost:
$C = 10(2x^2) + (6)(2)(2xh)+ (6)(2)(xh)$
$C = 20x^2 + 36xh$
$C = 20x^2 + (36x)(\frac{V}{2x^2})$
$C = 20x^2 +\frac{18V}{x}$
We can find the point where $C'(x) = 0$:
$C(x) = 20x^2 +\frac{18V}{x}$
$C'(x) = 40x -\frac{18V}{x^2} = 0$
$40x =\frac{18V}{x^2}$
$40x^3 =18V$
$x^3 =\frac{18V}{40}$
$x =\sqrt[3] {\frac{18V}{40}}$
$x =\sqrt[3] {\frac{(18)(10~m^3)}{40}}$
$x =\sqrt[3] {\frac{9~m^3}{2}}$
$x =\sqrt[3] {\frac{9}{2}}~m$
$x = 1.65~m$
When $0 \lt x \lt 1.65$, then $C'(x) \lt 0$
When $x \gt 1.65$, then $C'(x) \gt 0$
Thus, $x=1.65~m$ is the point where $C$ is a minimum.
We can find the minimum possible cost:
$C(1.65) = 20(1.65)^2 +\frac{(18)(10)}{1.65}$
$C(1.65) = 163.54$
The minimum possible cost of materials is $~~\$163.54$
