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# 求極值(偏微分

Find max &min Values of

F(x,y)=(x-4)^2+y^2

By the domain

D={(x,y),0≤x≤2,x^3≤y≤4x}

解max=68,min=10

### 1 Answer

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- densityLv 61 decade agoFavorite Answer
Firstly, we note that the domain of f is a compact set. Thus f attain its maximum and minimum. fx=2(x-4); fy =2y , Letting fx=0, fy=0. we have (x, y)=(4,0) be a critical point. fxx=2; fxy= 0 = fyx ; fyy=2 . The Hessian matrix H(x,y)=fxx．fyy- fxy． fyx so that H(4,0)= 4>0. Thus the minimum of f is f(4,0)=16. Next, we know that the extreme occur at the critical points or on its boundary. So we must test the boundary point (x, y)=(2,8),hence f(2,8)=68 is th maximum of f.

Source(s): myself

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