A grain silo has the shape of a right circular cylinder surmounted by a hemisphere. If the silo is to have a volume of 497π ft^3, determine the radius and height of the silo that requires the least amount of material to build. Hint: The volume of the silo isπ (r^2) (h) + 2/3 π r^3,and the surface area (including the floor) is π (3r^2 + 2rh). (Round your answers to one decimal place.)
π (r^2) (h) + 2/3 π r^3 =497 solve for h h= 497/pi r^2 - 2r/3 Surface area = 3pi r^2 + 2pi r h = 3pi r^2 + 2 pi r [ 497/pi r^2 - 2r/3] Surface area = 3 pi r^2 +994 /r - (4/3)pi r^2 dS/dr = 3pi (2r) -994/r^2 -(4/3) pi (2r) = 0 6pir--994/r^2 - (8/3)pi r = 0 6pi r^3 -994 -(8/3) pi r^3 = 0 (10/3) pi r^3 = 994 r^3 = (994)(3/10)(1/pi) =94.92 r= (94.92)^(1/3) r =4.5616 h= 497/pi r^2 - 2r/3 = 497/pi (4.5616)^2 - 2(4.56)/3 =4.56 d^2S/dr^2 =6pi +1988/r^3 -(8pi/3) = 10pi/3 +1988/r^3 > 0 for r=4.56 This shows the surface area has been minimized. r=4.6 h=4.6 (after rounding to 1 decimal place)
First note that in this problem, the height h refers only to the height of the cylindrical portion of the silo, not including the domed part (which would add another r ft to the height). The formulas given for the volume surface area include both portions of the silo. Begin by finding an expression of h in terms of r. V = pi(r^2)(h) + (2/3)(pi)(r^3) 497(pi) = pi(r^2)h + (2/3)(pi)(r^3) 497(pi) - (2/3)(pi)(r^3) = pi(r^2)h [497(pi) - (2/3)(pi)(r^3)] / (pi)(r^2) = h 497/r^2 - (2/3)r = h Now, using the formula for the surface area and substituting for h in terms of r, the function f(r) that calculates the amount of material needed based on the radius of the silo is: f(r) = (pi)(3r^2 + 2rh) f(r) = (pi)[3r^2 + 2r(497/r^2 - (2/3)r)] f(r) = (pi)[3r^2 + 994/r - (4/3)r^2] f(r) = (pi)( (5/3)r^2 + 994/r ) Use the First Derivative Test to find the value of r where f(r) takes on its minimal value. f '(r) = (pi)( (10/3)r - 994/r^2 ) 0 = (pi)( (10/3)r - 994/r^2 ) 0 = (10/3)r - 994/r^2 994/r^2 = (10/3)r 994 = (10/3)r^3 2982/10 = r^3 cbrt(2982/10) = r 6.6809 = r h = 497/r^2 - (2/3)r h = 497/6.6809^2 - (2/3)6.6809 h = 497/44.6346 - 4.4539 h = 11.1349 - 4.4539 h = 6.6710 Actually, there's a lot of rounding error in that value of h. If you use a calculator and don't round off, you'll see that the height h is the same as the radius r (accurate to 9 decimal places). Regardless, when rounded to one decimal place your answers are: r = 6.7 ft h = 6.7 ft (again, not including the domed portion of the silo; 13.4 ft with the dome). Edited to add: Cidyah beat me to the punch... but unfortunately, Cidyah did not calculate h in terms of r correctly... the total volume of the silo is 497(pi) according to the question. Everything else he/she did in terms of methodology is spot on; but unfortunately his/her final answers end up being incorrect.
