The surface area of a can is givin by the equation below \\(S_a=\\pi r^{2} + 2\\

Question

The surface area of a can is givin by the equation below

   (S_a=pi r^{2} + 2pi Rh+ pi R^{2} + pi(R +r)sqrt{(R-r)^{2}+(4.8-h)})   

Find the values that minimize the surface area with the following Volume Constraints

   (V=(pi (4.8-h)(R^{2}+Rr+r^{2})/3 )+ pi R^{2}h)   

Where

Please show your work. This is actually only part of a larger problem. So messing up here will ruin the rest of the work. I understand that it is simply rewriting the volume equation in terms of R or h. Inserting the new expression into the surface area equation, and then taking the derivative of the surface area equation. Lastly you set the differential equation equal to zero and solve for your one varible (R or h). However the derivative and algebra that follow is NASTY, I know you matlab pros could crunch this one out for me :)

Explanation / Answer

clc;

clear all;

syms R h

V=355.6;

r=2.215;

f=pi*((4.8-h)*(R^2+R*r+r^2)/3)+pi*R^2*h-V;

s=pi*r^2+2*pi*R*h+pi*R^2+pi*(R+h)*((R-r)^2+(4.8-h))^0.5;

ff=solve(f,h);

optimum_value=subs(s,h,ff);

diff_optimum_value=diff(optimum_value);

Radius=solve(diff_optimum_value);

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