The rectangle has dimensions: 2r and x.
Tne semicircle: 21⋅2⋅π⋅r=π⋅r .
p=2r+2x+π⋅rx=21(p−r(π+2))
The area of the window is to be a maximum:
S=x⋅2r+21π⋅r2==(p−r(π+2))⋅r+21π⋅r2==pr−r2(π+2)+21π⋅r2S′=p−2r(π+2)+πr=p−πr−4rp−πr−4r=0r(π+4)=pr=π+4p
If r∈(−∞,π+4p),S′>0 .
If r∈(π+4p,∞),S′<0 .
S(π+4p) - max.
r=π+4p .
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