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.
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
