Let f(x,y)=3x^3-9x^2+xy^2-y^2. a. Compute the gradient \nabla f(x,y) b. Find the critical points...

Question:

Let {eq}f(x,y)=3x^3-9x^2+xy^2-y^2. {/eq}

a. Compute the gradient {eq}\nabla f(x,y) {/eq}

b. Find the critical points of {eq}f {/eq} by solving {eq}\nabla f(x,y)=0, {/eq}

c. Use the second partial derivative test to classify each critical point as a local maximum, local minimum, or saddle point.

Hessian

The critical points of a function can be maximum, minimum or saddle point, a way to know the type of critical point is through the second derivative of the function and 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{then (a,b, f(a,b)) is a saddle point} \\ D(a,b) = 0 \Rightarrow \; \; \text{ the test is inconclusive} \\ {/eq}

Answer and Explanation:

We have the function

{eq}f(x,y)=3x^3-9x^2+xy^2-y^2 {/eq}

a.

{eq}\nabla f(x,y) =f_{x} i + f_{y} j \\ {/eq}

First partial derivatives:

{eq}f_x=9\,{x}^{2}+{y}^{2}-18\,x \\ {/eq}

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

{eq}\nabla f(x,y) =(9\,{x}^{2}+{y}^{2}-18\,x ) i + ( 2\,xy-2\,y ) j \\ {/eq}

b.

Now,

{eq}f_x=0\,\, \Rightarrow \, 9\,{x}^{2}+{y}^{2}-18\,x =0\\ f_y=0\,\, \Rightarrow \, 2\,xy-2\,y =0\\ {/eq}

The possible critical points are:

{eq}(0,0) \\ (2, 0) \\ (1, 3) \\ (1, -3) \\ {/eq}

c.

Second partial derivatives:

{eq}f_{xx}=18\,x-18 \\ f_{yy}=2\,x-2 \\ f_{yx}= 2\,y \\ {/eq}

Second derivative test for:

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

{eq}f_{xx} (0,0)= -18 \\ f_{yy} (0,0)= -2 \\ f_{xy} (0,0)= 0 \\ {/eq}

Hessian is given by:

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

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

{eq}\displaystyle D(0,0)= f_{xx}(0,0).f_{yy}(0,0)-[ f_{xy} (0,0) ]^{2} \\ \displaystyle D(0,0)= -18*-2-[0]^{2} \\ \displaystyle D(0,0)= 36 \\ {/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 (0,0) & -18 & -2 & 0 & 36 & Relative\, maximum \; & (0,0, 0) \\ (2,0) & 18 & 2 & 0 & 36 & Relative\, minimum \; & (2,0, -12) \\ (1, 3) & 0 & 0 & 6 & -36 & Saddle \, point \; & (1,3, -6) \\ (1, -3) & 0 & 0 & -6 & -36 & Saddle \, point \; & (1,-3, -6) \\ \\ \end{array} \\ {/eq}


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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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