A silo(base not included) is to be constructed in the form of cylinder surmounted by a hemisphere. The cost of the construction per square unit of surface area is 9 times for the hemisphere as it is for the cylindrical sidewall. determine the dimensions to be used if the volume is fixed at 10000 cubic units and the cost of construction is to be kept to a minimum. neglect the thickness of the silo and waste of the construction. a)what is the function of the silo C in terms of the radius rb)what is the interval of interest of the objective functionc)what is the radius of the cylindrical base(and of the hemisphere) d)the height of the cylindrical base thx
Let h and r be the height and radius of the *cylinder*. Thus the dome has radius r also. The volume is the vol of cylinder + vol half sphere = 10000= pr2h + (1/2) 4pr3/3, p = pi, notation etc... Here we can isolate h as a function of r; h = h(r) [do it!] The price of the total area is 2prh + 9* (1/2) 4pr2 = 2p( rh + 9r2 ), to be minimized. So drop the 2p for simplicity and replace your h = h(r): Minimize r*h(r) + 9r^2 and you have your answers!
A = 2πrh + 2πr^2, and if we take sidewall as unit cost C = 2πrh + 18πr^2, ....................(1) but we need this in terms of radius only, so we use V = πr^2h + (2/3)πr^3 2V/r = 2πrh + (4/3)πr^2 2πrh = 20,000r^(-1) - (4/3)πr^2 ...(2) C = 20,000/r + [18 - (4/3)]πr^2 C = 20,000r^(-1) + (50/3) πr^2 [This is (a)] As to what is the interval of interest of the objective function; it might mean what range of r might be used to graph C(r) to include the required minimum. C cannot be less than zero, and the cube root of 20000 is about 27, so you could use 0 < r < 25 dC/dr = (100/3)πr – 20000/r^2 = 0 for min r = (600/π)^(1/3) ~ 5.75882 units [This is (c), base and hemisphere] V/πr^2 = h + 2r/3 h = 1000/ πr^2 - 2r/3 ~ 92.141 units [This is (d)] Regards – Ian H
