D. X=10 all you have to do is pick an answer and plug it in
Let say radius is r
<span>its height is h </span>
<span>its lateral area = y </span>
<span>y = 2 pi r h </span>
<span>since the cylinder is inscribed in the sphere </span>
<span>So (2r )^2 + h^2 = 64 </span>
<span>then 4 (r^2) = 64 - h^2 </span>
<span>since y^2 = 4 (pi)^2 r^2 h^2 </span>
<span>then y^2 = (pi)^2 *h^2 * (64 -h^2) </span>
<span>y^2 = 64 (pi)^2 * h^2 - (pi)^2 * h^4 </span>
<span>2 y y' = 128 (pi)^2 * h - 4 (pi)^2 * h^3 </span>
<span>putting y' = 0 </span>
<span>4 (pi)^2 h ( 32 - h^2)=0 </span>
<span>ether h = 0 testing this value (changing of the sign of y' before and after ) y is minimum </span>
<span>or h = 4 sqrt(2) </span>
<span>testing this value (changing of the sign of y' before and after ) y is maximum </span>
<span>So the maximum value of y^2 = (pi)^2 *32 *( 64 - 32) </span>
<span>y^2 = (pi)^2 * (32)^2 </span>
<span>y = 32 (pi) square feet
hope this helps</span>
3(2x^3+3y^2) (2x^3 -3y^2)
