MATH 251 Lecture 14

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Maxima and Minima

From Calculus 1, critical points on a graph of $y=f(x)$ are given by $y^{\prime }=f^{\prime }(x)=0$.

For $z=f(x,y)$, we do the same thing, only in both the $x$ and $y$ direction.

$\mathrm{d}f(x_0,y_0)=0\{\begin{array}{l}{f}_x(x_0,y_0)=0\\ f_y(x_0,y_0)=0\end{array}$

Take the partial derivatives and set them both to 0.

Also remember to check the boundary of domain.

Example

Find $z=9-x^2-y^2$ on the domain $[-3,3]\times [-4,4]=\{(x,y)|-3\le x\le 3,-4\le y\le 4\}$

The domain of this function in the $xy$ plane is a rectangle.

Step 1: Find the Critical points

We need to find $\frac{\partial f}{\partial x}=0$

$\begin{array}{rl}{z}_x& =-2x=0\\ z_y& =-2y=0\\ \end{array}$

Solving this system of equations simultaneously is simple: $(x,y)=(0,0)$, so the critical point is at the origin.

Step 2: Check Boundaries

$y=4$, $x\in [-3,3]$

therefore, $f(x,4)=9-x^2-16=-x^2-7$. Take its derivative ($-2x$) to calculate any critical points on the boundary ($(x,y)=(0,4)$)

Do the same for the other three boundaries, and the critical points along the boundaries are:

(0,4), (0,−4), (3,0), (−3,0)

Also check the corners of the boundaries:

(3,4), (−3,4), (−3,−4), (3,−4)

Step 3: Check All values

1. $f(0,0)=9$
2. $f(\pm 3,0)=0$
3. $f(\pm 3,\pm 4)=-16$
4. $f(0,\pm 4)=-7$

Therefore, the global maximum is 9 @ (0,0) and the global minimum is -16 @ (±3, ±4)

Example

$z=xy\sin \left(\pi (x-y)\right)$ over the region bounded by $y\le \left|\frac{x}{2}\right|$, $x\ge 0$, and $x^2-y^2=1$

…

So $y=0$ or $\tan \left(\pi (x-y)\right)=-\pi x$
and $x=0$ or $\tan \left(\pi (x-y)\right)=-\pi y$

Check (0,0), (1,0), and solution of above simultaneously.