MATLAB: Answers given as equation

evaluating

Hello,

I have the equation;

% Radius = r
r = 10*u.ft;
SurfaceAreaOfASphere = (4*pi*r^2)

Returning an answer of;

400*pi*[ft]^2

But I want an answer without pi. In a sense evaluate the expression, any ideas?

Do you mean this?

vpa(SurfaceAreaOfASphere)
ans =
1256.6370614359172953850573533118*[ft]^2

hi Suman Koirla, try this :

The Volume is given by : V=(1/3)*r²*h, and the surface A=pi*r*sqrt(r²+h²)

for V=10 m^3, we search for r that minimizes the Surface , :

 Min(A)   ,  SUBject to  V=10

we have : h=3*V/pi*r² then : A=pi*r*sqrt(r²+90/pi²*r^4) .

Min(A) means the dA/dr=0=......=4*pi*r^3-180 /(2*sqrt(pi*r^4+90/r²))=0

Fast way to find R :

 syms r
 A=(pi^2*r^2+90/r^2)^1/2
 ezplot(A)
 S=subs(A,-6:0.1:6); % AXIS based on the first graph
 min(S)

1)so the minimum value for S=29.83 meter is R=1.89 ( FROM THE GRapH )

2) The corresponding value for h=3*10/(pi*1.89)=5.0525 meter .

Not possible. Since r and h are interdependent, "fmincon" must be able to vary both and thus predict them simultaneously.

Of course, if we assume that maximum volume will result if we use maximum surface area, we have

10 = 2*pi*r^2 + 2*pi*r*h

and we can solve for h:

h = 5/(pi*r) - r

and insert in the volume formula

max: pi*r^2*(5/(pi*r) - r)

But for bigger problems with more constraints, this does not seem the way to go.