A wire of 3 m length is to be used to make a circile and a square. How should the wire be distributed between two shapes in order to minimize the sum of the enclosed areas?

Can anyone write matlab code of this question?

A wire of 3 m length is to be used to make a circile and a square. How should the wire be distributed between two shapes in order to minimize the sum of the enclosed areas?

[madhan ravi]

you need to show what you have tried so far

[Busra Tabak]

syms a r

f=pi*r^2+a^2;

g=2*pi*r+4*a-3;

r1=solve(g==0,r);

f=subs(f,r,r1);

as=double(solve(diff(f,a)==0,a));

as=as(as>0);

rs=(3-4*as)/(2*pi);

fdp=diff(f,a,as);

fdp=double(subs(fdp,a,as));

if fdp>0

disp('min')

end

l1=2*pi*rs

l2=4*as

I wrote this and got 2 distributed shapes. Correct?

min

l1 =

1.3197

l2 =

1.6803

Answer 1

[Stephan]

Hi,

your code returns the correct result. Here is another way to write the same in symbolic form:

syms l1 l2 % declare symbolic variables
A_quad = (l1/4)^2; % area of square expressed with l1
A_circ = (l2/pi)^2*pi/4; % areas of circle expressed with l2
A = A_quad + A_circ; % sum of areas
A = subs(A,l2,l1-3); % using the constraint that l1 + l2 = 3 to eliminate l2
A_diff = diff(A,l1) == 0; % calculate first derivative and set it equal to zero
A_min = double(diff(lhs(A_diff),l1)) % check 2. derivative --> if positive it is a minimum
res = solve(A_diff,l1) % solve for l1 (analytic expression)
sol_l1 = double(res) % get a numeric value for the solution above

This returns:

A_min =

    0.2842

 
res =
 
12/(pi + 4)
 

sol_l1 =

    1.6803

Since A_min is positive you know it is a minimum. You can also check the result numeric within 2 lines of code:

% Check numeric
fun = @(l)((l(1)/4)^2 + (l(2)/pi)^2*pi/4);
num_sol_l1 = fmincon(fun,[3 0],[],[],[1 1],3)

The result is:

num_sol_l1 =

    1.6803    1.3197

Best regards

Stephan

[Busra Tabak]

Perfect thanks! I understood

[madhan ravi]

+1