Answer to Question #138341 in Calculus for Promise Omiponle

Question #138341
What are the local maxima, minima, and saddle points of the following function? What are the values of the function at those points? f(x, y) = (1 +xy)(x+y)
1
Expert's answer
2020-10-18T17:16:41-0400

Given, function is


f(x,y)=(1+xy)(x+y)=x+y+x2y+y2xf(x,y)=(1+xy)(x+y)=x+y+x^2y+y^2x

Now, critical points are determined by f(x,y)=0\nabla f(x,y)=0 , thus


f(x,y)=(1+2xy+y2,1+2xy+x2)=0    1+2xy+y2=0,1+2xy+x2=0    y2=x2    y=±x\nabla f(x,y)=(1+2xy+y^2,1+2xy+x^2)=0\\ \implies 1+2xy+y^2=0,1+2xy+x^2=0\\ \implies y^2=x^2\implies y=\pm x


Clearly, yxy\ne x as 1+3y20,yR1+3y^2\ne 0,\forall y\in \mathbb{R}

Thus, we conclude that y=x    1x2=0    x=±1y=-x\implies 1-x^2=0\implies x=\pm1

Therefore critical points are (1,-1) and (-1,1)

Now, we have to check the Hessian matrix for f at critical points to check whether above critical points are maxima, minima or saddle points.

Now,


Hf(x,y)=[2y2(x+y)2(x+y)2x]Hf(x,y)=\begin{bmatrix} 2y&2(x+y)\\ 2(x+y)&2x \end{bmatrix}

Now, clearly Hf(1,1)=[2002]&Hf(1,1)=[2002]Hf(1,-1)=\begin{bmatrix} -2&0\\0&2 \end{bmatrix}\&Hf(-1,1)=\begin{bmatrix} 2&0\\0&-2 \end{bmatrix}

Clearly, both Hf(1,1)&Hf(1,1)Hf(1,-1)\&Hf(-1,1) are diagonal hence their eigen values are ±2\pm 2

Hence, both Hf(1,1)&Hf(1,1)Hf(1,-1)\&Hf(-1,1) are semi positive definite ,thus (1,-1) and (-1,1) are saddle points.


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