Science:Math Exam Resources/Courses/MATH105/April 2011/Question 07

MATH105 April 2011

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[hide] Question 07
Find numbers $a$ and $b$ such that the function $f(x,y)=\ln(xy^{2})+bx^{2}+3axy$ has a critical point at $(1,3).$ Is this critical point a local maximum, a local minimum, or a saddle?

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[show] Hint
What do partial derivatives at critical points look like?

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[show] Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The derivative of $f$ with respect to $x$ is $f_{x}(x,y)=(1/x)+2bx+3ay$ . Also, the derivative of $f$ with respect to $y$ is $f_{y}(x,y)=(2/y)+3ax$ . For a critical point at $(1,3)$ , we require that $f_{x}(1,3)=f_{y}(1,3)=0$ . Using the above two equations, this gives us the following system to solve for $a$ and $b$ : $f_{x}(1,3)=(1/1)+2b(1)+3a(3)=1+2b+9a=0$ $f_{y}(1,3)=(2/3)+3a(1)=2/3+3a=0$ . Equation 2 tells us that $a=-2/9$ . Substituting this value for $a$ into equation 1 tells us that $1+2b+9(-2/9)=1+2b-2=2b-1=0$ . Therefore, we have that $b=1/2$ . This critical point is a saddle point. To show this, we solve for the second derivatives $f_{xx}$ , $f_{xy}$ and $f_{yy}$ . First, $f_{xx}(x,y)=-(1/x)^{2}+2b$ . Therefore, $f_{xx}(1,3)=-(1)+2(1/2)=0$ . Second, $f_{yy}(x,y)=-2(1/y)^{2}$ . Therefore, $f_{yy}(1,3)=-2/9$ Third, $f_{xy}(x,y)=3a$ . Therefore, $f_{xy}(1,3)=-2/3$ . We can classify the critical points by looking at the formula, $D=f_{xx}f_{yy}-(f_{xy})^{2}.$ If $D<0$ then the critical point is a saddle point. If $D>0$ then we look to the sign of $f_{xx}$ . If $f_{xx}>0$ then we have a local minimum and if $f_{xx}<0$ then we have a local maximum. For our problem, $D(1,3)=0\left(-{\frac {2}{9}}\right)-\left(-{\frac {2}{3}}\right)^{2}=-{\frac {4}{9}}<0.$ Therefore since $D<0$ the critical point is a saddle point. Advanced: Using the second partial derivatives of a function $f(x)$ we can define a general Hessian matrix as ${\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}}$ which for our specific example is $\left[{\begin{array}{cc}0&-2/3\\-2/3&-2/9\end{array}}\right]$ . The eigenvalues of this matrix are given by the solving the formula $(-\lambda )(-2/9-\lambda )-4/9=0$ for $\lambda$ . That is, $\lambda ^{2}+(2/9)\lambda -4/9=0$ . Solving gives us the answer $\lambda =(1/9)(\pm {\sqrt {37}}-1)$ . Because one eigenvalue is positive and one eigenvalue is negative, we conclude that the Hessian matrix is indefinite, and thus the point $(1,3)$ is a saddle point.