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

Explanation:

Step1: Expand the surface - area formula

Given $A = 2\pi r(r + h)=2\pi r^{2}+2\pi rh$. Rearrange it to the standard quadratic - form $ax^{2}+bx + c = 0$. So, $2\pi r^{2}+2\pi rh - A=0$, where $a = 2\pi$, $b = 2\pi h$, and $c=-A$.

Step2: Use the quadratic formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substituting $x = r$, $a = 2\pi$, $b = 2\pi h$, and $c=-A$ into the quadratic formula, we get $r=\frac{-2\pi h\pm\sqrt{(2\pi h)^{2}-4(2\pi)(-A)}}{2(2\pi)}$.

Step3: Simplify the expression

First, simplify the numerator: [ \begin{align*} \sqrt{(2\pi h)^{2}-4(2\pi)(-A)}&=\sqrt{4\pi^{2}h^{2}+8\pi A}\ &=2\sqrt{\pi^{2}h^{2} + 2\pi A} \end{align*} ] Then, $r=\frac{-2\pi h\pm2\sqrt{\pi^{2}h^{2}+2\pi A}}{4\pi}=\frac{-\pi h\pm\sqrt{\pi^{2}h^{2}+2\pi A}}{2\pi}$. Another way is to start from $A = 2\pi r(r + h)$ and solve for $r$ as follows: [ \begin{align*} A&=2\pi r^{2}+2\pi rh\ r^{2}+rh-\frac{A}{2\pi}&=0 \end{align*} ] Using the quadratic formula for $r^{2}+rh-\frac{A}{2\pi}=0$ (where $a = 1$, $b = h$, $c=-\frac{A}{2\pi}$), we have $r=\frac{-h\pm\sqrt{h^{2}+4\times\frac{A}{2\pi}}}{2}=\frac{-h\pm\sqrt{h^{2}+\frac{2A}{\pi}}}{2}=\sqrt{\frac{A}{2\pi}+\frac{h^{2}}{4}}-\frac{h}{2}$ (we take the positive root since $r>0$).

Answer:

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