An observatory is to be in the form of a right circular cylinder surmounted by a
ID: 3287522 • Letter: A
Question
An observatory is to be in the form of a right circular cylinder surmounted by a hemispherical dome. If the hemispherical dome costs 9 times as much per square ft as the cylindrical wall, what are the most economic dimensions for a volume of 10,000 cubic feet? The radius of the cylindrical base (and of the hemisphere) is ___ ft. ? round to the nearest tenthExplanation / Answer
Note that the surface area of the hemispherical dome is: S(dome) = 2?r^2, (This is just half of the surface area of a sphere, 4?r^2.) while the surface area of the cylindrical part is: S(cylinder) = ?r^2 + ?rh. (Note that this is not 2?r^2 + ?rh since the top part of the cylinder is covered.) Say that the cylindrical part costs $1 per square foot and the dome part costs $5 per square foot. The total cost is then: C = S(cylinder) + 5*S(dome) = ?r^2 + ?rh + 5(2?r^2) = 11?r^2 + ?rh. The volume of the observatory is: V(cylinder) + V(dome) = ?r^2h + (2/3)?r^3. Since the volume is 4000 ft^3, we have: ?r^2*h + (2/3)?r^3 = 4000. Solving this for h yields: h = [4000 - (2/3)?r^3]/(?r^2) = (12000 - 2?r^3)/(3?r^2). Substituting this into the cost function gives: C = 11?r^2 + ?rh = 11?r^2 + ?r[(12000 - 2?r^3)/(3?r^2)] = 11?r^2 + (12000 - 2?r^3)/(3r) = 11?r^2 + 4000/r - (2/3)?r^2 = (31/3)?r^2 + 4000/r. Differentiating yields: dC/dr = (62/3)?r - 4000/r^2. Setting this equal to zero gives: (62/3)?r - 4000/r^2 = 0 ==> r ? 3.94. Then: h = (12000 - 2?r^3)/(3?r^2) ? [12000 - 2?(3.94)^3]/[3?(3.94)^2] ? 79.39. Thus, the radius of the cylindrical base is 3.94 ft and the height is 79.39 ft.