Find the dimensions of a cylinder of volume 31( m^3 )of minimal cost if the top

Question

Find the dimensions of a cylinder of volume 31( m^3 )of minimal cost if the top and bottom are made of material that costs twice as much as the material for the side.

Explanation / Answer

volume= base area*height 31=pi r^2*h ===>h= 31/(pi*r^2) assume the side material cost is A the total cost will be C= pi*r^2*2A+2*pi*r*h*A =pi*r^2*2A+2*pi*r*A/(pi*r^2) c'=pi*2r*2A-2A/r^2 the optimum when c'=0 then pi*2r*2A -2A/r^2=0 if we multiply the equation by r^2, will get pi*2r^3*2A-2A=0 r=(1/(2pi))^(1/3)=0.542 m 2 A*B=144 V=(A-6)*(B-6)*3 V=(144/B-6)*(B-6)*3 =540-18B-2592/B d'v/dB= -18+2592/B^2 the optimum when the d'v=0 then -18+2592/B^2=0 B=12 and A= 12 the bases area will be 36 and the maximum volume 108 m^3

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