# Find all points where the given function has any local extrema. Give the values of any local...

## Question:

Find all points, if they exist, of local extrema on the given function. Give the values of any local extrema. Identify any saddle points.

{eq}\displaystyle\; f(x,y) = -6x^{2} - 6xy - 9y^{2} - 5x + 44y + 6 {/eq}

## Local Extrema of a Two Variable Function:

To categorize a critical point {eq}(a,b) {/eq} of a two variables function {eq}f(x,y) {/eq} depends on the following equation:

{eq}D = f_{xx} f_{yy} - \left( f_{xy} \right)^2 {/eq}

There are four cases which we will use to examine the local or relative extrema of this function and identify the saddle points:

Case 1:

If {eq}D(a,b) > 0 \, \, and \,\, f_{xy} (a,b) > 0 {/eq} then the function has local minimum at point {eq}(a,b) {/eq}

Case 2:

If {eq}D(a,b) > 0 \, \, and \,\, f_{xy} (a,b) < 0 {/eq} then the function has local maximum at point {eq}(a,b) {/eq}

Case 3:

If {eq}D(a,b) < 0 {/eq} then the saddle point is {eq}(a,b) {/eq}

Case 4:

If {eq}D(a,b) = 0 {/eq} then the function may be local maximum, local minimum or saddle point at {eq}(a,b) {/eq}

To initially find the coordinate (a, b), we must first equate the first partial differentials to zero and solve for x and y.

{eq}f(x,y) = -6x^2 - 6xy - 9y^2 - 5x + 44y + 6 {/eq}

The first order partial derivatives of this function are:

{eq}f_x = -12x -6y -5 \, \,\, and \,\,\, f_y = -6x - 18y + 44 {/eq}

The second order partial derivatives of this function are:

{eq}f_{xx} = -12 \,\,\, , \,\,\, f_{yy} = -18 \,\,\, and \,\,\, f_{xy} = -6 {/eq}

To find the critical point of this function, we have to set {eq}f_x = 0 \,\, and \,\, f_y = 0 {/eq}

Hence,

{eq}f_x = 0 \\ \Rightarrow -12x - 6y - 5 = 0 \\ \Rightarrow -12x = 6y + 5 \\ \Rightarrow x = - \frac{6y + 5}{12} \hspace{1 cm} \text{(Equation 1)} {/eq}

And,

{eq}f_y = 0 \\ \Rightarrow -6x - 18y + 44 = 0 \\ \Rightarrow -6 \left( - \frac{6y + 5}{12} \right) - 18y + 44 = 0 \hspace{1 cm} \left[ \because x = - \frac{6y + 5}{12} \right] \\ \Rightarrow \frac{6y +5}{2} - 18y + 44 = 0 \\ \Rightarrow 6y + 5 - 36y + 88 = 0 \\ \Rightarrow -30y + 93 = 0 \\ \Rightarrow -30y = - 93 \\ \Rightarrow y = \frac{93}{30} \\ \Rightarrow y = 3.1 {/eq}

Inserting {eq}y = 3.1 {/eq} into (Equation 1), we have:

{eq}x = - \frac{6(3.1) + 5}{12} \\ \Rightarrow x = - \frac{23.6}{12} \\ \Rightarrow x = - \frac{5.9}{3} {/eq}

So, the critical point is {eq}\left( - \frac{5.9}{3} , 3.1 \right) {/eq}

Now, in classifying the point, we will consider the following {eq}D {/eq} :

{eq}\begin{align*} D&= f_{xx} f_{yy} - \left( f_{xy} \right)^2 \\ &= (-12)(-18) - (-6)^2 \\ &= 216 - 36 \\ &= 180 \end{align*} {/eq}

At {eq}\left( x,y \right) = \left( - \frac{5.9}{3} , 3.1 \right) {/eq} :

{eq}D = 180 > 0 {/eq}

And:

{eq}f_{xy} = -6 < 0 {/eq}

Hence, we can say that the function has only one local minimum at the point {eq}\left( - \frac{5.9}{3} , 3.1 \right) {/eq} and there is no saddle point. The local maximum value of this function is:

{eq}\begin{align*} f( -\frac{5.9}{3} , 3.1 ) &= -6(- \frac{5.9}{3})^2 - 6( -\frac{5.9}{3})(3.1) - 9(3.1)^2 - 5(- \frac{5.9}{3}) + 44(3.1) + 6 \\ &= - \frac{69.62}{3} + 36.58 - 86.49 + \frac{29.5}{3} + 136.4 + 6 \\ &= - \frac{69.62}{3} + \frac{29.5}{3} + 92.49 \\ & \approx 79 \end{align*} {/eq}