What now Batman??? Optimization Problem Cont.....

Okay, found the algebra easier if I made the square "L-x" and let the circle be "x"

so

the sum of the two areas would be

Circle + Square = 2Pi(x/2Pi)(squared)+((L-x)/4)(squared)

I think that is right. So the next step should be to take a derivative, SO I will do the left hand side first, add the right hand later (this can get ugly looking quickly here):

Left hand side:

A'(x)=[(Pix(squared))'(4Pi(square))-Pix(squared)(4Pi(squared)]/16Pi(squared)

[(2xPi)(4Pi(squared))-0]/(16Pi(fourth))

8xPi(third)/16Pi(fourth)

x/2Pi for the left hand side

right hand now:

A'(x)=((L-x)/4)(Squared)

2(L-x)(-1)(16)-0/16

-2(L-x)/16

-(L-x)/8

Combined back:

x/2Pi + -L+x/8

(4+xPi-Lpi)/8Pi

so set top to zero

4+xPi-LPi=o

Stuck again, been at this for hours, I know I need to find both max and min, but the L is causing me great problems. The original problem does not address L or ask for it, could I use a simple value of 1 "one" for it, as the ratio of any cut of the wire would apply to all lengths. I need to walk away for a moment, bang my head against a solid object, multiple times (no I will not take a derivative to find v of that), and hope someone reads this a gives me the push in the right direction

Signed

Soon to have a headache