Question

find the critical point of the function ( f(x,y)=2e^{x}-3xe^{y} ).
the critical point is
use the second derivative test to determine whether it is
a. a local maximum
b. a saddle point
c. a local minimum
d. test inconclusive
submit answer next item
answers

Explanation:

Step1: Find first - order partial derivatives

The first - order partial derivatives of \(f(x,y)=2e^{x}-3xe^{y}\) are:
\(f_{x}=\frac{\partial f}{\partial x}=2e^{x}-3e^{y}\)
\(f_{y}=\frac{\partial f}{\partial y}=-3xe^{y}\)
Set \(f_{x}=0\) and \(f_{y}=0\).
From \(f_{y} = 0\), we have \(-3xe^{y}=0\). Since \(e^{y}\gt0\) for all \(y\in R\), then \(x = 0\).
Substitute \(x = 0\) into \(f_{x}=0\): \(2e^{0}-3e^{y}=0\), i.e., \(2 - 3e^{y}=0\). Solving for \(y\), we get \(e^{y}=\frac{2}{3}\), so \(y=\ln(\frac{2}{3})\).

Step2: Find second - order partial derivatives

\(f_{xx}=\frac{\partial^{2}f}{\partial x^{2}}=2e^{x}\)
\(f_{xy}=\frac{\partial^{2}f}{\partial x\partial y}=-3e^{y}\)
\(f_{yy}=\frac{\partial^{2}f}{\partial y^{2}}=-3xe^{y}\)

Step3: Calculate the discriminant \(D\)

The discriminant \(D=f_{xx}f_{yy}-(f_{xy})^{2}\).
At the point \((0,\ln(\frac{2}{3}))\):
\(f_{xx}(0,\ln(\frac{2}{3}))=2e^{0} = 2\)
\(f_{xy}(0,\ln(\frac{2}{3}))=-3e^{\ln(\frac{2}{3})}=-2\)
\(f_{yy}(0,\ln(\frac{2}{3}))=-3\times0\times e^{\ln(\frac{2}{3})}=0\)

\(D=(2)\times(0)-(- 2)^{2}=-4\lt0\)

Answer:

The critical point is \((0,\ln(\frac{2}{3}))\). Since \(D=-4\lt0\), the point is a saddle point. So the answer is B. A saddle point.