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ollegr [7]
3 years ago
9

Consider the following equation. f(x, y) = e−(x − a)2 − (y − b)2 (a) Find the critical points. (x, y) = a,b (b) Find a and b suc

h that the critical point is at (−3, 8). a = b = (c) For the values of a and b in part (b), is (−3, 8) a local maximum, local minimum, or a saddle point?
Mathematics
1 answer:
murzikaleks [220]3 years ago
8 0

a.

f(x,y)=e^{-(x-a)^2-(y-b)^2}\implies\begin{cases}f_x=-2(x-a)e^{-(x-a)^2-(y-b)^2}\\f_y=-2(y-b)e^{-(x-a)^2-(y-b)^2}\end{cases}

Critical points occur where f_x=f_y=0. The exponential factor is always positive, so we have

\begin{cases}-2(x-a)=0\\-2(y-b)=0\end{cases}\implies(x,y)=\boxed{(a,b)}

b. As the previous answer established, the critical point occurs at (-3, 8) if \boxed{a=-3} and \boxed{b=8}.

c. Check the determinant of the Hessian matrix of f(x,y):

\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}

The second-order derivatives are

f_{xx}=(-2+4(x-a)^2)e^{-(x-a)^2-(y-b)^2}

f_{xy}=4(x-a)(y-b)e^{-(x-a)^2-(y-b)^2}

f_{yx}=4(x-a)(y-b)e^{-(x-a)^2-(y-b)^2}

f_{yy}=(-2+4(y-b)^2)e^{-(x-a)^2-(y-b)^2}

so that the determinant of the Hessian is

\det\mathbf H(x,y)=f_{xx}f_{yy}-{f_{xy}}^2=\left((4(x-a)^2-2)(4(y-b)^2-2)-16(x-a)^2(y-b)^2\right)e^{-2(x-a)^2-2(y-b)^2}

\det\mathbf H(x,y)=(16(x-a)^2(y-b)^2-8(x-a)^2-8(y-b)^2)+4)e^{-2(x-a)^2-2(y-b)^2}

The sign of the determinant is unchanged by the exponential term so we can ignore it. For a=x=-3 and b=y=8, the remaining factor in the determinant has a value of 4, which is positive. At this point we also have

f_{xx}(-3,8;a=-3,b=8)=-2

which is negative, and this indicates that (-3, 8) is a local maximum.

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