Surface areas of solids

Solving for the dimensions of the cylinder:

let $y$ be the height and $r$ be the radius

By similar similar triangles, we have

$\dfrac{y}{6-r}=\dfrac{12}{6}=2$ $\implies$ $y=12-2r$

Find the volume of the right circular cylinder.

$V=\pi r^2y=\pi r^2(12-2r)$

Differenciate with respect to $r$ applying the product rule.

$\dfrac{dV}{dr}=\pi[r^2(-2)+(12-2r)(2r)]=\pi(-2r^2+24r-4r^2)=\pi(-6r^2+24r)$

For maximum volume, $\dfrac{dV}{dr}=0$ .

$\pi(-6r^2+24r)=0$ $\implies$ $6r^2=24r$ $\implies$ $r=4$

It follows that,

$y=4$

Solving for the surface area of the resulting solid:

The surface area of the resulting solid is equal to the lateral area of a frustum of a cone plus lateral area of the right circular cylinder plus area of the base. The lateral area of the frustum of a cone is $\frac{1}{2}$ multiplied by the sum of the circumference of the two bases multiplied by the slant height, that is $\frac{1}{2}(8\pi+12\pi)(\sqrt{20})=10\pi\sqrt{20}$ . The lateral area of the cylinder is circumference of the base multiplied by the height, that is $8\pi(4)=32\pi$ . The area of the base is area of the big circle minus area of the small circle, that is $\dfrac{\pi}{4}(12^2-8^2)=20\pi$ .

Summing up the three areas, we get

$\pi(10\sqrt{20}+52)$

Finally,

$a+b+c=10+20+52=$ $\boxed{\large\color{#D61F06}82}$