设f(x,y)=kx2+2kxy+y2在点(0,0)处取得极小值,求k的取值范围.

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问题 设f(x,y)=kx2+2kxy+y2在点(0,0)处取得极小值,求k的取值范围.

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答案由f(x,y)=kx2+2kxy+y2,可得fx’(x,y)一2kx+2ky,fyy’’(x,y)=2k,fy’(x,y)=2kx+2y,fyy’’(x,y)=2,fxy’’(x,y)=2k,于是,①若△=B2一AC=4k2一4k<0且A一2k>0,故0<k<1;②若△=B2一AC=4k2一4k=0,则k=0或k=1,当k=0时,f(x,y)=y2,由于f(x,0)≡0,于是点(0,0)非极小值点.当k=1时,f(x,y)=(x+y)2,由于f(x,一x)≡0,于是点(0,0)也非极小值点.综上所述,k的取值范围为(0,1).

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