The problem can be stated as a maximization/minimization.
Find maximum/minimum of
S(x,y) = 2pi x y+2 pi x^2S(x,y)=2πxy+2πx2
subjected to
x^2+y^2=r^2x2+y2=r2
y > 0y>0
Introducing a slack variable ss to transform the inequality restriction into an equality equivalent restriction we have
g_1(x,y) = x^2+y^2-r^2 = 0g1(x,y)=x2+y2−r2=0
g_2(x,y) = y-s^2=0g2(x,y)=y−s2=0
The lagrangian is
L(x,y,lambda_1,lambda_2,s) = S(x,y)+lambda_1 g_1(x,y) + lambda_2 g_2(x,y,s)L(x,y,λ1,λ2,s)=S(x,y)+λ1g1(x,y)+λ2g2(x,y,s)
being analytic, the stationary conditions are given by
grad L(x,y,lambda_1,lambda_2,s)=vec 0∇L(x,y,λ1,λ2,s)=→0
giving
{
(2 lambda_1 x + 4 pi x + 2 pi y = 0),
(lambda_2 + 2 pi x + 2 lambda_1 y = 0),
( x^2 + y^2 -r^2= 0),
(y -s^2= 0),
(2 lambda_2 s = 0)
:}
Solving for x,y,lambda_1,lambda_2,s we obtain
(x= -r, y= 0, lambda_1= -2pi, lambda_2= 2pi r, s= 0),
(x = r, y = 0, lambda_1= -2pi, lambda_2 = -2pi r, s = 0),
(x= -r/2 sqrt[2 -sqrt[2]], y=r/2 sqrt[2 + sqrt[2]], lambda_1 =(sqrt[2]-1)pi, lambda_2 = 0,
s ne 0),
(x=r/2 sqrt[2 + sqrt[2]], y= r/2 sqrt[2 - sqrt[2]], lambda_1 =-(sqrt[2]+1)pi, lambda_2 = 0,
s ne 0)
As we kow the active restriction is g_1(x,y) so, the stationary points qualification is made over
f_{g_1}(x)=2 pi x^2 + 2 pi x sqrt[r^2 - x^2]
having a minimum (d^2f_{g_1})/(dx^2)>0 at
(x= -r/2 sqrt[2 -sqrt[2]], y=r/2 sqrt[2 + sqrt[2]], lambda_1 =(sqrt[2]-1)pi, lambda_2 = 0,
s ne 0),
and maximum (d^2f_{g_1})/(dx^2)<0 at
(x=r/2 sqrt[2 + sqrt[2]], y= r/2 sqrt[2 - sqrt[2]], lambda_1 =-(sqrt[2]+1)pi, lambda_2 = 0,
s ne 0)
The maximum value being
S(r/2 sqrt[2 + sqrt[2]], r/2 sqrt[2 - sqrt[2]]) = (1 + sqrt[2]) pi r^2
If r = 1 then S(1/2 sqrt[2 + sqrt[2]], 1/2 sqrt[2 - sqrt[2]]) = (1 + sqrt[2]) pi