MATH 251 Lecture 18

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Quadratic Forms

$Q(x,y)=ax^{2}+2bxy+cy^{2}$

For the second derivative test, we replace $f(x,y)$ with its second order Taylor polynomial:

$f(x_{0}+\Delta x,y_{0}+\Delta y)\approx f(x_{0},y_{0})+{\frac {1}{2}}\left(f_{xx}(x_{0},y_{0})\cdot (\Delta x)^{2}+2f_{xy}(x_{0},y_{0})\Delta x\Delta y+f_{yy}(x_{0},y_{0})\cdot (\Delta y)^{2}\right)$

Classification

Examine the values of $Q(x,y)$ for $f(x,y)$ on the unit circle $x^{2}+y^{2}=1$. Find two relative extrema $\lambda _{1}$ and $\lambda _{2}$

if all positive

a paraboloid whose vertex represents a minimum

if all negative

a paraboloid whose vertex represents a maximum

if mixed

a saddle surface

if some value on degenerate

a parabolic cylinder representing a line of minimum or maximum values

Lagrange Multipliers

Useful for finding extrema along a curve or surface domain (just the problem we were discussing earlier).

Find relative extrema of $f(x,y)$ for $(x,y)$ constrained by $g(x,y)=C$.

$g(x,y)=C$ represents a level curve of the function $g$

The general idea is to follow the curve $g(x,y)=C$ along the direction of $\nabla f$ to find the maximum value. At all extrema points, $\nabla f$ and $\nabla g$ will be parallel. Since they are parallel, one will be a scalar multiple of the other.

We're left with the following 3 equations with unknown values $x$, $y$, and $\lambda$.

${\begin{cases}{\frac {\partial f}{\partial x}}=\lambda {\frac {\partial g}{\partial x}}\\{\frac {\partial f}{\partial y}}=\lambda {\frac {\partial g}{\partial y}}\\g(x,y)=C\end{cases}}$

Example

Going back to our previous line of thought, find the extrema of $2xy=f(x,y)=Q(x,y)$ on $g(x,y)=x^{2}+y^{2}=1$:

${\begin{aligned}\nabla Q=\left\langle 2y,2x\right\rangle \\\nabla g=\left\langle 2x,2y\right\rangle \end{aligned}}$

Thus our three equations are ${\begin{cases}2y=\lambda 2x\\2x=\lambda 2y\\x^{2}+y^{2}=1\end{cases}}$

Solving these equations gives $\lambda =\pm 1$ OR $x=0$.

• If $x=0$, then $y=0$, and (0,0) is not on our unit circle. We need a nonzero solution.
• If $\lambda =1$, then $x=y$, and the points are $\left({\tfrac {\sqrt {2}}{2}},{\tfrac {\sqrt {2}}{2}}\right),\ \left(-{\tfrac {\sqrt {2}}{2}},-{\tfrac {\sqrt {2}}{2}}\right)$
• If $\lambda =-1$, then $x=y$, and the points are $\left({\tfrac {\sqrt {2}}{2}},-{\tfrac {\sqrt {2}}{2}}\right),\ \left(-{\tfrac {\sqrt {2}}{2}},{\tfrac {\sqrt {2}}{2}}\right)$

Plugging all solutions for a respective $\lambda$ into $Q(x,y)$ gives the same $\lambda$ value... Coincidence? I think not!

For Quadratic forms, the Lagrange multipliers $\lambda$ are actually equal to the relative extrema. The values for $\lambda$ are called eigenvalues, and the characteristic polynomial we solved for $\lambda$ was $\lambda ^{2}-1=0$

In this case, $Q$ is mixed

Matrix Form

For $Q(x,y)=ax^{2}+2bxy+cy^{2}$,

$Q(x,y)=\left[A{\begin{pmatrix}x\\y\end{pmatrix}}\right]\cdot {\begin{pmatrix}x\\y\end{pmatrix}}$, where $A={\begin{pmatrix}a&b\\b&c\end{pmatrix}}$

$\nabla Q=\left\langle 2ax+2by,2bx+2cy\right\rangle$.

The Lagrange multiplier method states that $\nabla Q=\lambda \nabla g$

${\begin{aligned}2ax+2by&=\lambda 2x\\2bx+2cy&=\lambda 2y\\\end{aligned}}$

This can be simplified down to $A{\begin{pmatrix}x\\y\end{pmatrix}}=\lambda {\begin{pmatrix}x\\y\end{pmatrix}}$

For our previous problem $Q=2xy$, $A={\begin{pmatrix}0&1\\1&0\end{pmatrix}}$, which produces the same two equations for lambda with our constraint curve. <math>\begin{cases} y = \lambda x \\ x = \lambda y \\ x^2+y^2 = 1 \end{cases}