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

\[\begin{align*}{f_x} & = 0 : \,7 + 2y - 2x = 0\\ {f_y} & = 0 : - 8 + 2x + 3{y^2} = 0\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,x = 4 - \frac{3}{2}{y^2}\end{align*}\]

As shown above we solved the second equation for \(x\) and we can now plug this into the first equation as follows,

\[0 = 7 + 2y - 2\left( {4 - \frac{3}{2}{y^2}} \right) = 3{y^2} + 2y - 1 = \left( {3y - 1} \right)\left( {y + 1} \right)\,\,\,\,\,\,\, \to \,\,\,\,\,\,y = - 1,\,\,\,y = \frac{1}{3}\]

This gives two values of \(y\) which we can now plug back into either of our equations to find corresponding \(x\) values. Here is that work.

\(y = - 1 : x = 4 - \frac{3}{2}{\left( { - 1} \right)^2} = \frac{5}{2}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\frac{5}{2}, - 1} \right)\)

\(y = \frac{1}{3} : x = 4 - \frac{3}{2}{\left( {\frac{1}{3}} \right)^2} = \frac{{23}}{6}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\frac{{23}}{6},\frac{1}{3}} \right)\)

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 \(x = \frac{5}{2}\) above only because we assumed first that \(y = - 1\) and so that leads to the solution listed in first line above. Likewise, we only got \(x = \frac{{23}}{6}\) because we first assumed that \(y = \frac{1}{3}\) which leads to the second solution listed in the second line above. The points \(\left( {\frac{5}{2},\frac{1}{3}} \right)\) and \(\left( {\frac{{23}}{6}, - 1} \right)\) are NOT a solutions to this system as can be easily checked by plugging these points into the either of the equations 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 two critical points : \(\left( {\frac{5}{2}, - 1} \right),\,\,\left( {\frac{{23}}{6},\frac{1}{3}} \right)\).

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,

\[D\left( {x,y} \right) = {f_{x\,x}}{f_{y\,y}} - {\left[ {{f_{x\,y}}} \right]^2} = \left[ { - 2} \right]\left[ {6y} \right] - {\left[ 2 \right]^2} = - 12y - 4\] Show Step 4

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

\[\begin{align*} & \left( {\frac{5}{2}, - 1} \right) & : \hspace{0.25in} & D\left( {\frac{5}{2}, - 1} \right) = 8 > 0\,\,\,\,\,\,\,\,\,\,\,\,\,{f_{x\,x}}\left( {\frac{5}{2}, - 1} \right) = - 2 < 0 & \hspace{0.25in} & {\mbox{Relative Maximum}}\\ & \left( {\frac{{23}}{6},\frac{1}{3}} \right) & : \hspace{0.25in} & D\left( {\frac{{23}}{6},\frac{1}{3}} \right) = - 8 < 0 &\hspace{0.25in} & {\mbox{Saddle Point}}\end{align*}\]

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.