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obwod=40cm
obl x aby pole=Pmax
x- jeden pok prostokata
y - drugi bok prostokata
2x+2y=40------->y=20-x
P=x*y=x(20-x)=-x²+20x
f. kwadratowa ma maximum gdy a<0 i x1=-b/2a
x1=-20/(-2)=10
Odp. to jest kwadrat o boku 10cm