MATH 251 Lecture 18

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Quadratic Forms

$Q(x,y)=a{x}^2+2bxy+c{y}^2$

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

$f(x_0+\mathrm{\Delta }x,y_0+\mathrm{\Delta }y)\approx f(x_0,y_0)+\frac{1}{2}(f_{xx}(x_0,y_0)\cdot (\mathrm{\Delta }x{)}^2+2f_{xy}(x_0,y_0)\mathrm{\Delta }x\mathrm{\Delta }y+f_{yy}(x_0,y_0)\cdot (\mathrm{\Delta }y{)}^2)$

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 $\mathrm{\nabla }f$ to find the maximum value. At all extrema points, $\mathrm{\nabla }f$ and $\mathrm{\nabla }g$ 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 $\lambda$.

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

Example

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

$\begin{array}{r}\mathrm{\nabla }Q=⟨2y,2x⟩\\ \mathrm{\nabla }g=⟨2x,2y⟩\end{array}$

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

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 $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}),(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$
• If $\lambda =-1$, then $x=y$, and the points are $(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}),(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$

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 $Q(x,y)=a{x}^2+2bxy+c{y}^2$,

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

$\mathrm{\nabla }Q=⟨2ax+2by,2bx+2cy⟩$.

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

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

This can be simplified down to $A\left(\begin{array}{c}x\\ y\end{array}\right)=\lambda \left(\begin{array}{c}x\\ y\end{array}\right)$

For our previous problem $Q=2xy$, $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, 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}