Answer
The dimensions of the poster with the smallest area are $~~24~cm \times 36~cm$
Work Step by Step
Let $w$ be the width of the poster.
Let $h$ be the height of the poster.
We can use the area of the printed area to find an expression for the height $h$:
$(h-12)(w-8) = 384$
$h = \frac{384}{w-8}+12$
We can find an expression for the area of the poster:
$A = wh$
$A = (w)(\frac{384}{w-8}+12)$
$A = \frac{384~w}{w-8}+12w$
We can find $\frac{dA}{dw}$:
$\frac{dA}{dw} = \frac{384~(w-8)-384~w}{(w-8)^2}+12$
$\frac{dA}{dw} = \frac{-3072}{(w-8)^2}+\frac{12(w-8)^2}{(w-8)^2}$
$\frac{dA}{dw} = \frac{12(w-8)^2-3072}{(w-8)^2} = 0$
$12(w-8)^2-3072 = 0$
$(w-8)^2-256 = 0$
$(w^2-16w+64)-256 = 0$
$w^2-16w-192 = 0$
We can use the quadratic formula:
$w = \frac{16\pm \sqrt{(-16)^2-(4)(1)(-192)}}{2(1)}$
$w = \frac{16\pm \sqrt{1024}}{2}$
$w = \frac{16\pm 32}{2}$
$w = -8$ or $w = 24$
Since the width is a positive value, $w=24$
We can find $h$:
$h = \frac{384}{w-8}+12$
$h = \frac{384}{24-8}+12$
$h = \frac{384}{16}+12$
$h = 24+12$
$h = 36$
The dimensions of the poster with the smallest area are $~~24~cm \times 36~cm$
