the surface area of a cylinder, $a$, in terms of the radius of the base, $r$, and height, $h$, is given by…

Explanation:

Step1: Expand the formula

$A = 2\pi r(r + h)=2\pi r^{2}+2\pi rh$

Step2: Rearrange to quadratic - form

$2\pi r^{2}+2\pi rh - A=0$. This is a quadratic equation of the form $ax^{2}+bx + c = 0$ where $x = r$, $a = 2\pi$, $b = 2\pi h$, and $c=-A$.

Step3: Use the quadratic formula $r=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$r=\frac{-2\pi h\pm\sqrt{(2\pi h)^{2}-4\times(2\pi)\times(-A)}}{2\times(2\pi)}$ $r=\frac{-2\pi h\pm\sqrt{4\pi^{2}h^{2}+8\pi A}}{4\pi}$ $r=\frac{-2\pi h\pm2\sqrt{\pi^{2}h^{2} + 2\pi A}}{4\pi}$ $r=\frac{-\pi h\pm\sqrt{\pi^{2}h^{2}+2\pi A}}{2\pi}$ Another way: Start from $A = 2\pi r(r + h)$ $\frac{A}{2\pi}=r^{2}+rh$ Complete the square: $r^{2}+rh+\frac{h^{2}}{4}=\frac{A}{2\pi}+\frac{h^{2}}{4}$ $(r + \frac{h}{2})^{2}=\frac{A}{2\pi}+\frac{h^{2}}{4}$

Step4: Solve for $r$

$r+\frac{h}{2}=\pm\sqrt{\frac{A}{2\pi}+\frac{h^{2}}{4}}$ $r=\sqrt{\frac{A}{2\pi}+\frac{h^{2}}{4}}-\frac{h}{2}$ (we take the positive root since radius $r>0$)

Answer:

D. $r=\sqrt{\frac{A}{2\pi}+\frac{h^{2}}{4}}-\frac{h}{2}$