Answer
$9in\times18in$.
Work Step by Step
Step 1. Illustrate the scenario in the drawing shown.
Step 2. Assume the total width is $x$ in, and the total height is $y$ in; we have the total area $A=xy$
Step 3. Without the margins, the width of the printing area is $x-4$, and the height is $y-8$
Step 4. The total printing area is then $(x-4)(y-8)=50in^2$, which gives $y=\frac{50}{x-4}+8$
Step 5. The total area in step 2 becomes $A=\frac{50x}{x-4}+8x$
Step 6. To find the minimum in $A$, take its derivative to get $A'=\frac{50(x-4)-50x}{(x-4)^2}+8=-\frac{200}{(x-4)^2}+8$
Step 7. Let $A'=0$, we get $(x-4)^2=25, x=4\pm5$. As $x\gt0$, we have $x=9$ and $y=\frac{50}{9-4}+8=18$
Step 8. The above is a minimum because $A'$ changes signs from negative to positive when passing $x=9$, that is $..(-)..(9)..(+)..$ (concave up).
Step 9. We conclude that a minimum area of $A=9\times18=162in^2$ can be obtained when the total width is $9in$ and the total height is $18in$.
