We're talking about a silo, so we're talking about a cylinder with a hemisphere on top. So the volume of the silo part or the cylinder is π R-squared H and half of a sphere, or half of 4 thirds π R cubed. It's given to us in the problem that that equals 14,000. So if I solved for HI would get 14,000 / π R-squared -2 thirds R equaling H. Now we need to look at the surface area, the surface area of the cylinder part. If we thought about unwrapping that cylinder part, the surface area would be the circumference of the base 2π R times the height, so 2π RH the hemisphere. We want half of a full sphere or half of four Pi R-squared in the directions. It tells us that the hemisphere is 9 times the cost as the cylindrical side walls. So we're going to take that 2π RH and multiply by one, and we're going to take that 1/2 of four Pi R-squared and multiply by 9 because it's nine times as much. So we're going to get the surface area equaling 2π RH plus 18 Pi R-squared. If we take and substitute the H we found a minute ago, we get 14,000 Pi r ^2 -, 2/3 R substituted in for the H. If we multiply by the 2π R to those two terms, we get 28,000 / r -, 4 thirds Pi r ^2 + 18 Pi r ^2. If we combine our Pi R-squared portions, we get 50 / 3 Pi R squareds. We're going to take the first derivative of surface area -28,000 / r ^2 + 100 / 3 Pi R, setting the surface area equal to 0, the derivative of the surface area equaling 0, and solving for R. So we're going to take the one term to the other side and get all the Rs on one side, everything on the other. So that's going to reduce to r ^3 being 840 / π or the cube root of 840 / π, which is approximately 6.4. If we then go back to the H, we get H equaling 14,000 / π. We are going to take that R and square it. So the cube root of 840 / π ^2 I did not simplify or I did not use the rounding error. I used the exact -2 thirds, the cube root of 840 / π and I got 103.1 feet for the height of the cylinder. So 6.4 feet for the radius and 103.1 feet for the height.