16.2 Absolute minima and maxima

What is an endpoint of a circle?

false

Interpret the height $\sqrt{x^2+y^2}$ geometrically.

The minimum height is zero at $(0,0,0)$. The derivatives $z_x$ and $z_y$ do not exist there. The maximum height is $\sqrt{2}$ at $(\pm 1,\pm 1,\sqrt{2})$. There $z_x$ and $z_y$ exist but are not zero — those points would not be the highest points if it were not for the restriction $|x|,|y|\le 1$.

Check the boundary of the square as well as critical points inside the square.

$\text{min}=0\qquad \text{max}=\frac{2}{3\sqrt{3}}\approx0.385$

(a)

(b) The maximum and minimum values of $h(x,y)$ in $x^2+y^2\le 1$ are $3$ (at $(0,1)$) and $-3$ (at $(0,-1)$), respectively.

There are five places to check: the interior and four boundaries.

The minimum is $-2$ and the maximum is $6$.

$6-2\sqrt{5}$

Since the region is a triangle, your boundary will have three separate parts to check.

(a) (0,0) and (3,0) and (0,3) are saddle points
(1,1) is a local min

(b) The minimum is $-1$ at $(1,1)$ and the maximum is $80$ at $(4,4)$.

(a) $(1,1)$ is a saddle point and $(2,4)$ is a local min

(b) The min and max are $\frac{19}{27}$ and $5$, respectively.

There are two boundary lines. You'll want to find their intersections.

(a) $(0,0)$, $(6,0)$, $(0,3)$ are saddle points and $(2,1)$ is a local min

(b) The maximum value is $0$ and the minimum value is $4(4\sqrt{2}-6) \approx -1.37$.

Plugging in the boundaries should be quite easy if you choose your variables wisely

The coldest temperture is $-0.391$ and the coldest point is $(0,2)$.

When you see “classify critical points," think “second derivative test."

(a) $(0,-5)$ is a saddle point

(b) The smallest value of $g$ is $0$ at $(0,0)$ and the largest value is $21$ at $(\pm 2\sqrt{3},-1)$.

Suppose that the bends are made a distance $x$ from the ends of the fence and that the bends are through an angle $\theta$. Draw a sketch of the enclosure and figure out its area, as a function of $x$ and $\theta$.

$\frac{2500}{\sqrt{3}}$

Suppose that the box has side lengths $x$, $y$ and $z$.

The box has dimensions $(2V)^{1/3}\times(2V)^{1/3}\times 2^{-2/3}V^{1/3}$.

(a) The maximum and minimum values of $T(x,y)$ in $x^2+y^2\le 4$ are $20$ (at $(0,0)$) and $4$ (at $(\pm 2,0)$), respectively.

(b) $(0,2)$

If $(x,y,z)$ is on the plane, then you know $z=5-2x-y$. So, you can write $x^2 y^2 z$ as a function of only $x$ and $y$ by eliminating $z$.

The minimum value is $0$ on

The maximum value is $4$ at $(1,2,1)$.

(a) $x=1$, $y=\half$, $f\big(1,\half\big)=6$ (b) local minimum

(c) As $x$ or $y$ tends to infinity (with the other at least zero), $2x+4y$ tends to $+\infty$. As $(x,y)$ tends to any point on the first quadrant part of the $x$- and $y$–axes, $\frac{1}{xy}$ tends to $+\infty$.
Hence as $x$ or $y$ tends to the boundary of the first quadrant (counting infinity as part of the boundary), $f(x,y)$ tends to $+\infty$. As a result $\big(1,\half\big)$ is a global (and not just local) minimum for $f$ in the first quadrant. Hence $f(x,y)\ge f\big(1,\half\big)=6$ for all $x,y>0$.

The answer will be piecewise, depending on what exactly $a$ is.

If $a<\frac12$, then the closest point is the origin. If $a\ge\frac12$, then the closest points are the level curve where $z=a-\frac12$.

Instead of maximizing the total profit function, maximize the profit functions of each type of paper.

Profit is (revenue) minus (costs). If Ayan and Pipe work separately, then each seller only sees the cost and revenue from the lemonade that they themselves sold.

To find how much each seller will sell when they are working separately, find out which values of $q_A$ and $q_P$ end up with both individual profit functions being maximized.

To find out how much they'll sell when they're working together, use your assumption from part (c) to make the solving smoother.