this is the question:A silo in the shape of a cylinder is required to hold 8600m^3 of wheat. a) find an equation for the surface area of the silo in terms of the base radius b) find the minimum surface area requried to hold this amount of what, to the nearest square metre.I get part a) but i dont get part b) because i dont know how to derive the equation to get the min surface area.the equation is:2πr^2 + 17200/rplease tell me how to derive it step by step, i know how to derive things but the 17200 over the r is the part i dont getany help is greatly appreciated thank you :)
Differentiate the formula with respect to r. Rewrite the second term with a negative exponent and use the power rule, if you don't like the quotient rule: 2πr^2 + 17200r^-1 2*2πr^1 + 17200(-1*r^-2) 4πr - 17200r^-2 4πr - 17200/(r^2) Set it equal to zero and solve for r: 4πr - 17200/(r^2) = 0 4πr = -17200/(r^2) 4πr^3 = -17200 r^3 = -4300/π = about -1368.7 r = about -11.10, which makes no sense... Are you sure that formula is right for the surface area? It should be 2pi*r^2 + 2pi*r*h.
S=2πr^2 + 17200/r=2πr^2 + 17200(r)^-1 Differentiate dS/dr = 4πr -17200(r)^-2 dS/dr = 4πr -17200/r^2 For Max/Min put dS/dr =0 4πr -17200/r^2=0 4πr^3 =17200 r^3 =17200/4π =1368.7325105903 taking cube root r =11.1 meter S = 2π(11.1)^2 + 17200/(11.1)=774.6+1549.6=2324.2 cm^2..........Ans
