Minimizing total poster area given fixed printed area and margins

Since $xy = 96$, solve for $y$ to get $y = 96/x$. Substitute into the total area formula $A(x) = (x+2)(y+4) = (x+2)\left(\dfrac{96}{x}+4\right)$. Expanding gives $A(x) = 96 + 4x + \dfrac{192}{x} + 8 = 4x + \dfrac{192}{x} + 104$. The domain is $x>0$ since $x$ is a physical length.

Differentiate: $A'(x) = 4 - \dfrac{192}{x^2}$. Setting $A'(x)=0$ gives $x^2 = 48$, so $x = \sqrt{48} = 4\sqrt{3}$ (taking the positive root). Since $A''(x) = 384/x^3 > 0$ for $x>0$, this critical point is a minimum. Then $y = 96/x = 96/(4\sqrt3) = 24/\sqrt3 = 8\sqrt3$, and the minimum area is $A = 4(4\sqrt3) + 192/(4\sqrt3) + 104 = 16\sqrt3 + 16\sqrt3 + 104 = 32\sqrt3 + 104 \approx 159.4\text{ in}^2$.