Math, asked by dhasarasathy, 3 months ago

examine for extreme values x3+y3-3x-12y+20

Answers

Answered by riddhisingh305

2

Step-by-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

Answered by moinmohamed184

0

Answer:

what sorry I didn't understood hmm

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