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### Section 14.3 : Relative Minimums and Maximums

4. Find and classify all the critical points of the following function.

$f\left( {x,y} \right) = 3{y^3} - {x^2}{y^2} + 8{y^2} + 4{x^2} - 20y$

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Start Solution

We’re going to need a bunch of derivatives for this problem so let’s get those taken care of first.

$\begin{array}{c}{f_x} = - 2x{y^2} + 8x\hspace{0.75in}{f_y} = 9{y^2} - 2{x^2}y + 16y - 20\\ {f_{x\,x}} = - 2{y^2} + 8\hspace{0.5in}{f_{x\,y}} = - 4xy\hspace{0.5in}{f_{y\,y}} = 18y - 2{x^2} + 16\end{array}$ Show Step 2

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

\begin{align*}{f_x} & = 0 : \,\,\, - 2x{y^2} + 8x = 2x\left( {4 - {y^2}} \right) = 0\hspace{0.25in} \to \hspace{0.25in}y = \pm 2\,\,\,{\mbox{or}}\,\,\,x = 0\\ {f_y} & = 0 : \,\,\,\,9{y^2} - 2{x^2}y + 16y - 20 = 0\end{align*}

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

$$y = - 2 : 4{x^2} - 16 = 0\,\,\,\,\, \to \,\,\,\,\,x = \pm 2\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {2, - 2} \right)\,\,\,\,{\mbox{and}}\,\,\,\left( { - 2, - 2} \right)$$

$$y = 2 : - 4{x^2} + 48 = 0\,\,\,\,\, \to \,\,\,\,\,x = \pm 2\sqrt 3 \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {2\sqrt 3 ,2} \right)\,\,\,\,{\mbox{and}}\,\,\,\left( { - 2\sqrt 3 ,2} \right)$$

$$x = 0 : 9{y^2} + 16y - 20 = 0\,\,\,\,\, \to \,\,\,\,\,y = \frac{{ - 16 \pm \sqrt {976} }}{{18}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {0,\frac{{ - 16 - \sqrt {976} }}{{18}}} \right)\,\,\,\,{\mbox{and}}\,\,\,\,\left( {0,\frac{{ - 16 + \sqrt {976} }}{{18}}} \right)$$

As we noted in the first two problems in this section be careful to only write down the actual solutions as found in the above work. Do not just write down all possible combinations of $$x$$ and $$y$$ from each of the three lines above. If you do that for this problem you will end up with a large number of points that are not critical points.

Also, do not get excited about the “mess” (i.e. roots) involved in some of the critical points. They will be a fact of life with these problems on occasion.

So, in summary, this function has the following six critical points.

$\left( { - 2, - 2} \right),\,\,\left( {2, - 2} \right),\,\,\left( {2\sqrt 3 ,2} \right),\,\,\left( { - 2\sqrt 3 ,2} \right),\,\,\,\left( {0,\frac{{ - 16 - \sqrt {976} }}{{18}}} \right),\,\,\,\left( {0,\frac{{ - 16 + \sqrt {976} }}{{18}}} \right)$.

Show Step 3

Next, we’ll need the following,

\begin{align*}D\left( {x,y} \right) & = {f_{x\,x}}{f_{y\,y}} - {\left[ {{f_{x\,y}}} \right]^2}\\ & = \left[ { - 2{y^2} + 8} \right]\left[ {18y - 2{x^2} + 16} \right] - {\left[ { - 4xy} \right]^2}\\ & = \left[ { - 2{y^2} + 8} \right]\left[ {18y - 2{x^2} + 16} \right] - 16{x^2}{y^2}\end{align*} Show Step 4

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

\begin{align*} & \left( { - 2, - 2} \right) & : \hspace{0.25in} & D\left( { - 2, - 2} \right) = - 256 < 0 & \hspace{0.25in} & {\mbox{Saddle Point}}\\ & \left( {2, - 2} \right) & : \hspace{0.25in} & D\left( {2, - 2} \right) = - 256 < 0 & \hspace{0.25in} & {\mbox{Saddle Point}}\\ & \left( { - 2\sqrt 3 ,2} \right) & : \hspace{0.25in} & D\left( { - 2\sqrt 3 ,2} \right) = - 768 < 0 & \hspace{0.25in} & {\mbox{Saddle Point}}\\ & \left( {2\sqrt 3 ,2} \right) & : \hspace{0.25in} & D\left( {2\sqrt 3 ,2} \right) = - 768 < 0 & \hspace{0.25in} & {\mbox{Saddle Point}}\\ & \left( {0,\frac{{ - 16 - \sqrt {976} }}{{18}}} \right) & : \hspace{0.25in} & D\left( {0,\frac{{ - 16 - \sqrt {976} }}{{18}}} \right) = 180.4 > 0\,\,\,\,\,\,{f_{x\,x}}\left( {0,\frac{{ - 16 - \sqrt {976} }}{{18}}} \right) = - 5.8 < 0& \hspace{0.25in} & {\mbox{Relative Maximum}}\\ & \left( {0,\frac{{ - 16 + \sqrt {976} }}{{18}}} \right) & : \hspace{0.25in} & D\left( {0,\frac{{ - 16 + \sqrt {976} }}{{18}}} \right) = 205.1 > 0\,\,\,\,\,\,{f_{x\,x}}\left( {0,\frac{{ - 16 + \sqrt {976} }}{{18}}} \right) = 6.6 > 0 & \hspace{0.25in} & {\mbox{Relative Minimum}}\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.