# Find the critical points of F(x,y) = \frac{1}{3} x ^3 + x y ^2 - 25x - 4 y ^2 and identify each...

## Question:

Find the critical points of {eq}F(x,y) = \frac{1}{3} x ^3 + x y ^2 - 25x - 4 y ^2 {/eq}

and identify each as a maximum , minimum or saddle point

{eq}f(x) = x ^ 2 + 2 y {/eq}

## Critical Points:

The critical points of a function can be found with the Hessian {eq}D(a,b) {/eq}

If {eq}D(a,b) > 0 \; \text{&} \; f_{xx} > 0 \; \Rightarrow \; \; \text{ f has a relative minimum point at (a,b, f(a,b))} \\ D(a,b) > 0 \; \text{&} \; f_{xx} < 0 \; \Rightarrow \; \; \text{ f has a relative maximum point at (a,b, f(a,b))} \\ D(a,b) < 0 \; \; \Rightarrow \; \; \text{ (a,b, f(a,b)) is a saddle point} \\ D(a,b) = 0 \Rightarrow \; \; \text{ the test is inconclusive} \\ {/eq}

1.) We have the function:

{eq}F(x,y) = \frac{1}{3} x ^3 + x y ^2 - 25x - 4 y ^2 \\ {/eq}

First partial derivatives:

{eq}f_x={x}^{2}+{y}^{2}-25 \\ {/eq}

{eq}f_y=2\,xy-8\,y \\ {/eq}

Second partial derivatives:

{eq}f_{xx}=2\,x \\ f_{yy}= 2\,x-8 \\ f_{yx}= 2\,y \\ {/eq}

Now:

{eq}f_x=0\,\, \Rightarrow \, {x}^{2}+{y}^{2}-25 =0\\ f_y=0\,\, \Rightarrow \, 2\,xy-8\,y =0\\ {/eq}

The possible critical points are:

{eq}(5,0)\\ (-5,0)\\ (4,3)\\ (4, -3)\\ {/eq}

Second derivative test for:

Point: {eq}(5,0)\\ {/eq}

{eq}f_{xx} (5,0)=2\,(5)=10\\ f_{yy} (5,0)= 2\,(5)-8=2\\ f_{xy} (5,0)=2\,(0)=0\\ {/eq}

The Hessian is given by:

{eq}\displaystyle D(a,b)= f_{xx}(a,b).f_{yy}(a,b)-[ f_{xy} (a,b) ]^{2} \\ {/eq}

So, the Hessian for {eq}(5,0) {/eq} is:

{eq}\displaystyle D(5,0)= f_{xx}(5,0).f_{yy}(5,0)-[ f_{xy} (5,0) ]^{2} \\ \displaystyle D(5,0)=10*2-[0]^{2}\\ \displaystyle D(5,0)=20\\ {/eq}

{eq}\begin{array} \; \; \text{(a,b)} \; & { \ f_{xx} (a,b) } & { f_{yy} (a,b) } & f_{xy} (a,b) & D(a,b) & Conclusion & (a,b, f(a,b)) \\ \hline (5,0) & 10 & 2 & 0 & 4 & Relative\, minimum \; & (5,0, -{\frac {250}{3}}) \\ (-5,0) & -10 & -18 & 0 & 180 & Relative\, maximum \; & (-5,0, {\frac {250}{3}}) \\ (4,3) & 8 & 0 & 6 & -36 & Saddle \, point \; & (4, 3, -{\frac {236}{3}}) \\ (4,-3) & 8 & 0 & -6 & -36 & Saddle \, point \; & (4, -3, -{\frac {236}{3}}) \\ \end{array} \\ {/eq}

2) We have the function:

{eq}f(x, y) = x ^ 2 + 2 y \\ {/eq}

First partial derivatives:

{eq}f_x=2\,x \\ {/eq}

{eq}f_y=2 \\ {/eq}

Second partial derivatives:

{eq}f_{xx}= 2 \\ f_{yy}= 0 \\ f_{yx}=0 \\ {/eq}

The function has no critical points; therefore, it has no maximum and minimum values.