Find x,y Coordinates of Stationary Point: 2x^2-2xy+y^2+2x+5

[MarkFL]

Here is the question:

Find the (x; y) coordinates of the stationary point of: z(x,y) = 2x^2 - 2xy + y^2 + 2x + 5 and find the natu?


Find the (x; y) coordinates of the stationary point of:

z(x,y) = 2x^2 - 2xy + y^2 + 2x + 5

and find the nature of the stationary point.

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Krazy G,

We are given the function:

\(\displaystyle z(x,y)=2x^2-2xy+y^2+2x+5\)

First, we want to find the critical points by equating the first partials to zero:

\(\displaystyle z_x(x,y)=4x-2y+2=0\)

\(\displaystyle z_y(x,y)=-2x+2y=0\)

The second equation implies $y=x$, and substitution for $y$ into the first equation yields:

\(\displaystyle x=-1\)

and so the critical point is:

\(\displaystyle (x,y)=(-1,-1)\)

Now, to determine the nature of this critical point we may utilize the second partials test for relative extrema.

\(\displaystyle D(x,y)=z_{xx}(x,y)z_{yy}(x,y)-\left[z_{xy}(x,y) \right]^2=4\cdot2-(-2)^2=4\)

Since \(\displaystyle z_{xx}(x,y)=4>0\) and \(\displaystyle D(x,y)=4>0\), then we conclude that the critical value is the global minimum. Hence:

\(\displaystyle z_{\min}=z(-1,-1)=4\)