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 such 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?

a.[tex]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}[/tex]Critical points occur where [tex]f_x=f_y=0[/tex]. The exponential factor is always positive, so we have[tex]\begin{cases}-2(x-a)=0\\-2(y-b)=0\end{cases}\implies(x,y)=\boxed{(a,b)}[/tex]b. As the previous answer established, the critical point occurs at (-3, 8) if [tex]\boxed{a=-3}[/tex] and [tex]\boxed{b=8}[/tex].c. Check the determinant of the Hessian matrix of [tex]f(x,y)[/tex]:[tex]\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}[/tex]The second-order derivatives are[tex]f_{xx}=(-2+4(x-a)^2)e^{-(x-a)^2-(y-b)^2}[/tex][tex]f_{xy}=4(x-a)(y-b)e^{-(x-a)^2-(y-b)^2}[/tex][tex]f_{yx}=4(x-a)(y-b)e^{-(x-a)^2-(y-b)^2}[/tex][tex]f_{yy}=(-2+4(y-b)^2)e^{-(x-a)^2-(y-b)^2}[/tex]so that the determinant of the Hessian is[tex]\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}[/tex][tex]\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}[/tex]The sign of the determinant is unchanged by the exponential term so we can ignore it. For [tex]a=x=-3[/tex] and [tex]b=y=8[/tex], the remaining factor in the determinant has a value of 4, which is positive. At this point we also have[tex]f_{xx}(-3,8;a=-3,b=8)=-2[/tex]which is negative, and this indicates that (-3, 8) is a local maximum.