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If you have any doubts in the solution pleaseaskme in comments  f(x,y)=-5x^(2)-5y^(2)+3x+3y-5domain: x^(2)+y^(2) <= 9.Now domain can be divided into two domain(i) x^(2)+y^(2) < 9.(ii) x^(2)+y^(2)=9.on x^(2)+y^(2) < 9.Now critical point given By ((del f)/(del x),(del f)/(del y))=(0,0){:[=>quad(-10 x+3","-10 y+3)=(0","0)],[=>quad x=3//10","quad y=3//10]:}**(x,y)=(3//10,3//10)dots (1)on x^(2)+y^(2)=9,Now g(x,y):x^(2)+y^(2)-9=0By Langrage multipiers, quad h(x,y,lambda)=f(x,y)+lambda g(x,y)Now h(x,y,lambda)=-5x^(2)-5y^(2)+3x+3y-5+lambda(x^(2)+y^(2)-9)Now {:[(del h)/(del x)=0=>-10 x+2lambda x+3,=0],[(del h)/(del y)=0,=>-10 y+2lambda y+3=0]}quad x=y{:[:'x=y=>x^(2)+x^(2)=9quad=>quad x=+-(3)/(sqrt2)],[=>(x","y)=((3)/(sqrt2),(3)/(sqrt2))quad&quad(x","y)=(-(3)/(sqrt2),-(3)/(sqrt2))]:}Now f((3)/(10),(3)/(10))=-5((9)/(100))-5((9)/(100))+(9)/(10)+(9)/(10)-5f((3)/(10),(3)/(10))=(-45-45+90+90-500)/(100)=-(410)/(100)=-(41)/(10)=-4.1Now f((3)/(sqrt2),(3)/(sqrt2))=9sqrt2-50quad~~-37.2720f(-(3)/(sqrt2),-(3)/(sqrt2))=-50-9sqrt2~~-62.7279". "So absolute extrema;{:[" abosolute maximun "=-4.1quad" at "((3)/(10),(3)/(10))],[" absolute minimum "=-62.728" at "(-(3)/(sqrt2),-(3)/(sqrt2)).]:}Answer** (2) quad f(x,y)=-x^(2)-xy+2y^(2)-x-2y+3domain 1 <= x <= 9,1 <= y <= 10domain can be divided into these parts(i) 1 < x < 9,quad1 < y < 10(ii) x=1,quad1 <= y <= 10(iii) x=9,quad1 <= y ... See the full answer