find the critical point of the function ( f(x,y)=2e^{x}-3xe^{y} ).\nthe critical point is\nuse the second…

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.