MythAsian From: US Posts: 42 Votes: 84 |
A grain silo has the shape of a right circular cylindersurmounted by a hemisphere . If the silo is to have acapacity of 504Pift^3 find the radius and height of the silothat requires the least amount of material to construct. Hint: The volume of the silo is Pir^2h +(2/3)Pir^3,and the surface area (including the floor) isPi(3r^2+2rh) |

TinySupa2 From: US Posts: 41 Votes: 82 |
Volume of the silo =504π ft^3 Volume of the silo =V= πr^2h + (2/3)πr^3 = πr^2 [ h +(2/3)r ] Surface of the silo =S = π [ 3r^2 + 2rh ] = πr [ 3r +2h ] radius = r height = h from the expression of the volume we canexpress h by: h = (V/ πr^2) - (2/3) r then replace in the expression of S S = πr [ 3r + 2h ]= πr [ 3r +2( (V/ πr^2) - (2/3) r) ] = πr [ 3r + (2V/ πr^2) -(4/3) r ] = πr [ (2V/ πr^2) + (5/3) r ] = (2V/r) + (5π/3) r^2 Let's differentiate S in terms of r dS/dr = (- 2V)r^(-2)+ (10π/3) r when dS/dr =0 (10π/3) r = ( 2V)r^(-2)----> (10π/3) r^3 = ( 2V) -----> r^3 = 6V/10π = 3V /5π ----> r = (3V / 5π) ^(1/3) V = 504π ft^3; ---> r = 6.712 ft replace r in the previous expression ofh h = (V/ πr^2) - (2/3) r = [504π /π(6.712)2] -(2/3). 6.712 = 11.187 - 4.475 =6.712 ft the least amount of material toconstruct is achieved when r = h = 6.712 ft Hope this helps |

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