Question 927974
I think you have a typo. Both {{{r}}} and {{{x}}} are parts of a sum of lengths that is the perimeter, {{{P}}} , so they cannot be {{{P(pi+4)>P}}} .
Here is your Norman window:
{{{drawing(240,440,-1.2,1.2,-3.2,1.2,
rectangle(-1,0,1,-3),locate(-0.1,0,2r),locate(-0.98,-1.4,x),
locate(0.9,-1.4,x),locate(-0.1,-2.84,2r),
arc(0,0,2,2,180,360),locate(0.25,0.433,r),
arrow(0,0,0.5,0.866),arrow(0.5,0.866,0,0)
)}}} {{{P=2x+2r+pi*r}}}--->{{{P=2x+(2+pi)r}}} and {{{area=2xr+pi*r^2/2}}}
If we have a fixed perimeter, the maximum area will be obtained when
{{{x=r=P/(pi+4)}}} , and here is my proof:
{{{P=2x+(2+pi)r}}}<--->{{{P-(2+pi)r=2x}}}<--->{{{x=P/2-(2+pi)r/2}}}
Substituting into {{{area=2xr+pi*r^2/2}}} , we get
{{{area=(P-(2+pi)r)r+pi*r^2/2}}}
{{{area=Pr-(2+pi)r^2+pi*r^2/2}}}
{{{area=Pr-2r^2-pi*r^2+(pi/2)*r^2}}}
{{{area=Pr-2r^2-(pi/2)*r^2}}}
{{{area=Pr-(2+pi/2)*r^2}}}
{{{area=-((4+pi)/2)*r^2+Pr}}}
{{{area}}} is a quadratic function of {{{r}}}
A quadratic function like {{{f(x)-ax^2+bx+c}}} has a maximum if the leading coefficient (the {{{a}}} in the term with the square) is negative.
The maximum happens when the variable has the value {{{x=-b/2a}}}
In this case, the maximum happens when
{{{r=-P/(2(-(4+pi)/2)))=P/(4+pi)=P/(pi+4)}}}
Substituting {{{r=P/(pi+4)}}} into {{{P-(2+pi)r=2x}}} , we get
{{{P-(2+pi)(P/(pi+4))=2x}}}
{{{P-(2+pi)P/(pi+4)=2x}}}
{{{P(1-(2+pi)/(pi+4))=2x}}}
{{{P(pi+4-(2+pi))/(pi+4)=2x}}}
{{{P(pi+4-2-pi)/(pi+4)=2x}}}
{{{2P/(pi+4)=2x}}}--->{{{x=P/(pi+4)}}}