## Calculus: Early Transcendentals 8th Edition

The dimensions of the poster with the smallest area are $~~24~cm \times 36~cm$
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 $~~24~cm \times 36~cm$