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 $y'=f'(x)=0$.

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

$\mathrm {d} f\left(x_{0},\,y_{0}\right)=0{\begin{cases}f_{x}(x_{0},y_{0})=0\\f_{y}(x_{0},y_{0})=0\end{cases}}$

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 $\left[-3,3\right]\times \left[-4,4\right]=\left\{(x,y)~|~-3\leq x\leq 3,\ -4\leq y\leq 4\right\}$

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

Step 1: Find the Critical points

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

${\begin{aligned}z_{x}&=-2x=0\\z_{y}&=-2y=0\\\end{aligned}}$

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

Step 2: Check Boundaries

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

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

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

…

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

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