Now, let’s find the critical points for this problem. That means solving the following system.

\[\begin{align*}{f_x} & = 0 : 2\left( {y - 2} \right)x = 0\hspace{0.25in} \to \hspace{0.25in}y = 2\,\,\,{\mbox{or}}\,\,\,x = 0\\ {f_y} & = 0 : {x^2} - 2y = 0\end{align*}\]

As shown above we have two possible options from the first equation. We can plug each into the second equation to get the critical points for the equation.

\(y = 2 : {x^2} - 4 = 0\,\,\,\,\, \to \,\,\,\,\,x = \pm 2\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {2,2} \right)\,\,\,\,{\mbox{and}}\,\,\,\left( { - 2,2} \right)\)

\(x = 0 : - 2y = 0\,\,\,\,\, \to \,\,\,\,\,y = 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {0,0} \right)\)

Be careful in writing down the solution to this system of equations. One of the biggest mistakes students make here is to just write down all possible combinations of \(x\) and \(y\) values they get. That is not how these types of systems are solved!

We got \(x = \pm 2\) above only because we assumed first that \(y = 2\) and so that leads to the two solutions listed in that first line above. Likewise, we only got \(y = 0\) because we first assumed that \(x = 0\) which leads to the third solution listed above in the second line. The points \(\left( {0,2} \right)\), \(\left( { - 2,0} \right)\) and \(\left( {2,0} \right)\) are NOT solutions to this system as can be easily checked by plugging them into the second equation in the system.

So, do not just “mix and match” all possible values of \(x\) and \(y\) into points and call them all solutions. This will often lead to points that are not solutions to the system of equations. You need to always keep in mind what assumptions you had to make in order to get certain \(x\) or \(y\) values in the solution process and only match those values up with the assumption you had to make.

So, in summary, this function has three critical points : \(\left( {0,0} \right),\,\,\left( { - 2,2} \right),\,\,\left( {2,2} \right)\).

Also, before proceeding with the next step we should note that there are multiple ways to solve this system. The process you used may not be the same as the one we used here. However, regardless of the process used to solve the system, the solutions should always be the same.

Show Step 3

Next, we’ll need the following,

\[D\left( {x,y} \right) = {f_{x\,x}}{f_{y\,y}} - {\left[ {{f_{x\,y}}} \right]^2} = \left[ {2\left( {y - 2} \right)} \right]\left[ { - 2} \right] - {\left[ {2x} \right]^2} = - 4\left( {y - 2} \right) - 4{x^2}\] Show Step 4

With \(D\left( {x,y} \right)\) we can now classify each of the critical points as follows.

\[\begin{align*} & \left( {0,0} \right) & : \hspace{0.25in} & D\left( {0,0} \right) = 8 > 0 \hspace{0.25in}{f_{x\,x}}\left( {0,0} \right) = - 4 < 0 & \hspace{0.25in} & {\mbox{Relative Maximum}}\\ & \left( { - 2,2} \right) & :\hspace{0.25in} & D\left( { - 2,2} \right) = - 16 < 0 & \hspace{0.25in} & {\mbox{Saddle Point}}\\ & \left( {2,2} \right) & :\hspace{0.25in} & D\left( {2,2} \right) = - 16 < 0 & \hspace{0.25in} & {\mbox{Saddle Point}}\end{align*}\]

Don’t forget to check the value of \({f_{x\,x}}\) when \(D\) is positive so we can get the correct classification (i.e. maximum or minimum) and also recall that for negative \(D\) we don’t need the second check as we know the critical point will be a saddle point.