Note: A silo is a right circular cylinder surmounted by a hemisphere.Volume was given to be 504π ft^3My brain is fried trying to figure out this problem. I already have the primary equation as A=2πrh+2πr^2 and secondary equation as 504π=π(r^2)h+(2/3)π(r^3).I need help on how to successfully substitute the secondary equation into the primary equation and solving it for zero.Thanks for any help you can provide.
hehehehehe I am supposed to solve the same problem for an exam lol. Well I have also been struggling with it. I just found this online First, Solve for Area (what you are trying to maximize) in terms of one variable…..r looks the easiest: 504(pi)=(pi)r^2h+2/3(pi)r^3, h=504/(r^2)-2r/3 So A = (pi)(3r^2+rh)=pi*(3r^2 + 504/r – 2r^2/3) = 7(pi)r^2 + 504/r At the maximum dA/dr = 0, so we set the derivative = to zero and solve: dA/dr = d/dr(7(pi)r^2 + 504/r) = 14(pi)r/3-504/r^2=0 After multiplying both sides by 3r^2: r^3 - 108 = 0 r = 3*2^(2/3) Back substitute to get the height: h=504/(r^2)-2r/3 h=12*2^(2/3) But i am not sure it is right, so i will probably go to talk to my teacher tomorrow. if i do I will get online and give you the right answer. I hope it helps!!
PI r^2 h = 3456 h = 3456 / PI r^2 S=2PI r^2 + 2PI r h S = 2PI r^2 + 2 PI r [ 3456 / PI r^2] S = 2 PI r^2 + 6812 / r differentiate with appreciate to r dS/dr = 4PI r -6812 / r^2 = 0 4 PI r = 6812 / r^2 4 PI r^3 = 6812 PI r^3 = 1703 r^3 = 1703 / PI r = [1703 / PI] ^(a million/3) r =8.1537 h = 3456 / PI r^2 = sixteen.5468 dS^2/dr^2 > 0 while r=8.15; This verfies the exterior section has been minimized.
