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Silo calculus question--is this possible to solve?

A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit of surface area is 10 times as great for the hemisphere as it is for the cylindrical sidewall. I need to determine the dimensions to be used if the volume is fixed 12000 cubic units and the cost of construction is to be kept to a minimum. I also can neglect the thickness of the silo and waste in construction.I need to find the radius of the cylindrical base and hemisphereand the height of the cylidrical base.

Answer:

no soz
First, you need to get a function for the volume. Let r= radius of cylinder and hemisphere and h= height of cylinder v= total volume = 12000 The dome has a volume of (2(pi)r^3)/3 The cylinder volume is (pi)r^2*h v= (pi)r^2 * h + (2/3)(pi)r^3 = 12000 Solve to get h in terms of r leaving: h = 12000/((pi)r^2) - (2/3)r Now make a formula for the cost (c): c = 10 (area of hemisphere) + (area of cylinder) = 10 (2(pi)r^2) + 2(pi)rh = 20(pi)r^2 + 2(pi)r(12000/((pi)r^2) - (2/3)r) = (56/3)(pi)r^2 + 24000/r Take the derivative and solve for when dc/dr = 0 to find the the min. dc/dr = (112/3)(pi)r - 24000/r^2 = 0 r = 5.8928 and h = 106.07 I hope you can fill in the missing steps.
Uh yea. Volume of hemisphere is just 1/2 that of a sphere so: Vh = (1/2)[(4/3)pi*R^3] = (2/3)pi*R^3 where R is the radius of the sphere Volume of the cylinder is just the base area times the height: Area of base = pi*R^2 Height of cylinder = H Vc = [pi*R^2]*H Total volume = V = 12000 V = Vc + Vh = [(2/3)pi*R^3] + [pi*R^2]*H = 12000 Cost for construction. Cost for cylindrial portion = C per unit area Cost of hemisphere = Ch = 10C Area of hemisphere is ne half the surface area of a sphere so: Ah = (1/2)4*pi*R^2 = 2*pi*R^2 Area of cylinder (just the side, don't need top and bottom): Ac = 2*pi*R*H Total cost = M = Ah*Ch + Ac*C M = (2*pi*R^2)(10C) + (2*pi*R*H)C And you want to minimize M the total cost. Use the volume equation to solve for H: [(2/3)pi*R^3] + [pi*R^2]*H = 12000 H = {12000 - [(2/3)pi*R^3]}/[pi*R^2] Put this into the equation for M and you will have an equation with just R as a variable (C is a constant and it doesn't really matter what it is except insofar as giving you the actual cost). Since we don't care what the total cost is I will just set C=1 so we can forget about it. M = (2*pi*R^2)(10) + 2*pi*R*{12000 - [(2/3)pi*R^3]}/[pi*R^2] M = (2*pi*R^2)(10) + 2*{12000 - [(2/3)pi*R^3]}/[R] M = (2*pi*R^2)(10) + 2*{12000/(R) - [(2/3)*pi*R^2]} dM/dR = 40*pi*R + 2*{-12000/(R^2) - [(4/3)*pi*R]} Set this to 0 for the minimum 40*pi*R + 2*{-12000/(R^2) - [(4/3)*pi*R]} = 0 40*pi*R^3 - 24000 - (8/3))*pi*R^3 = 0 pi*R^3 - 600 - (1/15))*pi*R^3 = 0 R^3[pi - (1/15)*pi] = 600 R^3 = 600 / [pi(1 - 1/15)] R = 5.89 Vh = (2/3)*pi*R^3 = (2/3)*pi*(5.89)^3 Vh = 428.57 Vc = 12000 - Vh = 11571.43 Vc = [pi*R^2]*H = [pi*(5.89)^2]H H = 11571.43/[pi*5.89)^2] H = 106.17 So the radius is 5.89 ahd the height of the base if 106.17
Silo Base

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