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Examine for extreme values x3+y3-3x-12y+20 |
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Answer» -step EXPLANATION:(x,y)=x3+y3−3x−12y+20 ∇f(x,y)=(3x2−3,3y2−12)=(0,0) So 3x2=3⟺x=±1 3y2=12⟺y=±2 (fxxfxyfyxfyy)=(6x006y) So at (1,2):D>0 and FXX>0 |
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