Matematyka
tosia1993
2017-06-25 02:23:17
Znajdź ekstremum funkcji: f (x,y)=3x³ + 3x²y - y³- 15x
Odpowiedź
slawek112
2017-06-25 07:03:41

[latex]f(x,y)=3x^3+3x^2y-y^3-15x[/latex] [latex]cfrac{partial f}{partial x}(x,y)=9x^2+6xy-15[/latex] [latex]cfrac{partial f}{partial y}(x,y)=3x^2-3y^2[/latex] [latex]cfrac{partial f}{partial y}(x,y)=0[/latex] [latex]3x^2-3y^2=0[/latex] [latex]3(x-y)(x+y)=0[/latex] [latex]y=x lor y=-x[/latex] [latex]y=xRightarrowcfrac{partial f}{partial x}(x,y)=9x^2+6x^2-15=15x^2-15=15(x-1)(x+1)[/latex] [latex]cfrac{partial f}{partial x}(x,y)=0Leftrightarrow{x=1lor x=-1}[/latex] [latex]y=-xRightarrowcfrac{partial f}{partial x}(x,y)=9x^2-6x^2-15=3x^2-15=3(x-sqrt{5})(x+sqrt{5}[/latex] [latex]cfrac{partial f}{partial x}(x,y)=0Leftrightarrow{x=sqrt{5}lor x=-sqrt{5}}[/latex] Punkty, w których funkcja może mieć ekstremum to: [latex](1,1),(-1,-1),(sqrt{5},-sqrt{5}),(-sqrt{5},sqrt{5})[/latex] [latex]cfrac{partial^2f}{partial x^2}(x,y)=18x+6y[/latex] [latex]cfrac{partial^2f}{partial xpartial y}(x,y)=6x[/latex] [latex]cfrac{partial^2f}{partial y^2}(x,y)=-6y[/latex] [latex]W(x,y)=left|egin{array}{cc}18x+6y & 6x\ 6x & -6yend{array} ight|[/latex] [latex]W(1,1)=left|egin{array}{cc}24 & 6\ 6 & -6end{array} ight|=-144-36=-180<0[/latex] [latex]W(-1,-1)=left|egin{array}{cc}-24 & -6\ -6 & 6end{array} ight|=-144-36=-180<0[/latex] [latex]W(sqrt{5},-sqrt{5})=left|egin{array}{cc}12sqrt{5} & 6sqrt{5}\ 6sqrt{5} & 6sqrt{5}end{array} ight|=360-180=180>0[/latex] [latex]W(-sqrt{5},sqrt{5})=left|egin{array}{cc}-12sqrt{5} & -6sqrt{5}\ -6sqrt{5} & -6sqrt{5}end{array} ight|=360-180=180>0[/latex] W punkcie [latex](sqrt{5},-sqrt{5})[/latex] funkcja ma minimum lokalne równe [latex]-10sqrt{5}[/latex]. W punkcie [latex](-sqrt{5},sqrt{5})[/latex] funkcja ma maksimum lokalne równe [latex]10sqrt{5}[/latex].

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