Answer to Question #247307 in Calculus for Susan

Question

Given the function f(x,y) = e^x ( x^2 - y^2) find all critical points and classify each as a local maximum, local minimum, or neither.

Answer 1

Solution.

Find the partial derivatives:

$\frac{\partial f}{\partial x}=2xe^x+(x^2-y^2)e^x,$

$\frac{\partial f}{\partial y}=-2ye^x.$

Let's solve the system of equations$2xe^x+(x^2-y^2)e^x=0,\newline -2ye^x=0.$

From the second equation $y=0.$ Substitute $y=0$ in the first equation of the system and find x:

$(2x+x^2)e^x=0,\newline 2x+x^2=0,\newline x(2+x)=0,\newline x_1=0, x_2=-2.$

So we have two critical points (0,0) and (-2,0).

Find the partial derivatives of the second order:

$B=\frac{\partial ^2f}{\partial x\partial y}=-2ye^x,$

$A=\frac{\partial ^2f}{\partial x^2}=4xe^x+(x^2-y^2)e^x+2e^x,$

$C=\frac{\partial ^2f}{\partial y^2}=-2e^x.$

Let us calculate the value of these second-order partial derivatives at the critical points.

Calculate the values ​​for point (0; 0):

$A=2, B=0, C=-2.$

Calculate AC - B ² = -4 <0, then there is no global extremum.

Calculate the values ​​for point (-2; 0):

$A=-\frac{2}{e^2},B=0, C=-\frac{2}{e^2}.$

Calculate AC - B ² = 4 / e ⁴ > 0 and A <0, then at the point (-2; 0) there is a local maximum and $f_{max}$ (-2; 0) = 4 / e ²

^{Answer. Two critical points (0,0) and (-2,0). Point (-2,0) is a point of local maximum.}