Answer
the dimensions are $24$ $cm$ and $36$ $cm$
Work Step by Step
Let $x$ and $y$ be the dimensions of the poster.
$xy$ = $384$
$y$ = $\frac{384}{x}$
The total area is
$A(x)$ = $(8+x)\left(12+\frac{384}{x}\right)$ = $12\left(40+x+\frac{256}{x}\right)$
$A'(x)$ = $12\left(1-\frac{256}{x^{2}}\right)$
$A'(x)$ = $0$
$12\left(1-\frac{256}{x^{2}}\right)$ = $0$
$x$ = $16$
There is an absolute minimum when $x$ = $16$ since
$A'(x)$ $\lt$ $0$ for $0$ $\lt$ $x$ $\lt$ $16$
$A'(x)$ $\gt$ $0$ for $x$ $\gt$ $16$
When $x$ = $16$, $y$ = $\frac{384}{16}$ = $24$
so the dimensions are $24$ $cm$ and $36$ $cm$
