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Mama L [17]
3 years ago
7

Find the local maximum and minimum values and saddle point(s) of the function.

Mathematics
1 answer:
iogann1982 [59]3 years ago
4 0
f(x,y)=9e^y(y^2-x^2)
\nabla f=\left\langle-18xe^y,9ye^y(2+y)\right\rangle

Critical points occur where the gradient is zero. This is guaranteed whenever x=0 and either y=0 or y=-2.

The Hessian matrix for this function looks like

H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}-18e^y&-18xe^y\\-18xe^y&9e^y(2-x^2+4y+y^2)\end{bmatrix}

and has determinant

|H(x,y)|=-162e^{2y}(2+x^2+4y+y^2)

Maxima occur whenever the determinant is positive and f_{xx}. Minima occur whenever both the determinant and f_{xx} are positive. Saddle points occur whenever the determinant is negative.

At (0,0), you have a saddle point since the determinant reduces to -324, so (0,0,0) is the saddle point.

At (0,-2), the determinant is \dfrac{324}{e^4}>0 and f_{xx}(0,-2)=-\dfrac{18}{e^2}, so \left(0,-2,\dfrac{36}{e^2}\right) is a local maximum.

No other critical points remain, so you're done.
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