One of the stationary point of the function f(x, y) = x++y" - 2x + 4xy – 2y is
(A) (V2.-V2) (B) (2.-2) (C) (12.2) (D) (-2, 2)
Answers
Answer:
More generally you can just use the Hessian at the critical points. We have
fxxfxyfyy=12x2−4,=4,=12y2−4.
So at the critical points P1,2=±(2–√,2–√) we have
Hf(P1,2)=(12⋅2−44412⋅2−4).
The principal minors Δ1=20, Δ2=202−42=384 are thus both positive. Meaning that the Hessian form is positive definite, and we have local minima.
On the other hand at the remaining critical point P3=(0,0) the Hessian
Hf(P3)=(12⋅0−44412⋅0−4)
is negative semidefinite as the determinant =Δ2=0.
This is immediately obvious from the quadratic Taylor polynomial at P3 as well, because
T2,P3(x,y)=−2x2−2y2+4xy=−2(x−y)2.
The nature of this critical point is thus left in the dark by a study of the Hessian alone. Let's do some testing suggested by the form of the quadratic Taylor polynomial. If x=y=h, then
f(x,y)=f(h,h)=2h4−2(h−h)2=2h4>0.
On the other hand
f(h,−h)=2h4−2(h−(−h))2=−8h2+2h4<0
when h is sufficiently close to zero.
This means that P3 is not a local extremal point.
Michael Rozenberg already handled the global minima/maxima.
Answer:
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