3fdc0a…: “As of a fair ocean”

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Consider the function $F(x,y) = x^3+9x^2 + y^3-9y +1$ , which is defined for all $(x,y) \in \R ^2$ .

The function $F(x,y)$ has $\answer {4}$ critical points over $\R ^2$ .

This function relative extrema over $\R ^2$ .
This function absolute extrema over $\R ^2$ .

The $x$ -values for the critical points are $x=\answer {-6}$ and $x=\answer {0}$ (type the smaller $x$ -value first) and the $y$ -values for the critical points are $y=\answer {-\sqrt {3}}$ and $y= \answer {\sqrt {3}}$ (type the smaller $y$ -value first).

Now, let $R = \left \{ (x,y) \in \R ^2 ~ \bigg | ~ x^2 \leq y \leq 4 \right \}$ .

Select all of the following that are critical points for $F(x,y)$ over $\R ^2$ .

$(-6,-\sqrt {3})$ $(-6,\sqrt {3})$ $(0,-\sqrt {3})$ $(0,\sqrt {3})$

Select all of the following that are critical points for $F(x,y)$ that lie in $R$ .

$(-6,-\sqrt {3})$ $(-6,\sqrt {3})$ $(0,-\sqrt {3})$ $(0,\sqrt {3})$

The function $F(x,y)$ any absolute extrema over $R$ because:

all functions that have critical points have absolute extrema. no functions that have critical points have absolute extrema. all continuous functions have absolute extrema over any subset of $\R ^2$ . all continuous functions have absolute extrema over any bounded subset of $\R ^2$ . all continuous functions have absolute extrema over any closed subset of $\R ^2$ . all continuous functions have absolute extrema over any closed and bounded subset of $\R ^2$ .

The absolute extrema could occur at the critical point in $R$ or along the boundary.

We first note that at the critical point $\left (0,\sqrt {3}\right )$ , we have $F\left (0,\sqrt {3}\right ) = \answer {-6 \sqrt {3}+1}$ .

To analyze around the boundary, first sketch $R$ , and click on the Hint below to verify if your sketch is correct.

The region $R$ is shown below, along with the critical points of $F(x,y)$ .

$\bullet$ Along the boundary $y=x^2$ for $-2 \leq x \leq 2$ , we have

$F(x,y) = F(x,x^2) = \answer { x^3 + x^6 +1 }$

Moving forward, let’s call the function above $f(x)$ (i.e. define $f(x) := F(x,x^2)$ ). The critical points here occur when $x=-\sqrt [3]{\answer {2}}$ , $x=\answer {0}$ , and $x=\sqrt [3]{\answer {2}}$ .

The critical points will occur when $f'(x) = \ddx \left [ x^3 + x^6 +1 \right ] =0$ .

Checking the value of $F(x,y)$ at each of these as well as at the endpoints gives the following values.

$f\left (-2\right ) = \answer {57}$
$f\left (-\sqrt [3]{2}\right ) = \answer {3}$
$f\left (0\right ) = \answer {1}$
$f\left (\sqrt [3]{2} \right ) = \answer {7}$
$f\left (2 \right ) = \answer {73}$

$\bullet$ Along the boundary $y=4$ for $-2 \leq x \leq 2$ , we have

$F(x,y) = F(x,4) = \answer { x^3+9x^2 + 29 }$

Moving forward, let’s call the function above $g(x)$ (i.e. define $g(x) := F(x,4)$ ). The critical points here occur when $x=\answer {0}$ .

The critical points will occur when $g'(x) = \ddx \left [ x^3+9x^2 + 29 \right ] =0$ . Note this means that $x=0$ or $x=-3$ , the latter of which is not along the boundary since we need $-2 \leq x \leq 2$ .

Checking the value of $F(x,y)$ at each of these as well as at the endpoints gives the following values.

$g\left (-2\right ) = \answer {57}$
$g\left (0 \right ) = \answer {29}$
$g\left (2 \right ) = \answer {73}$

Note that $f(-2) = g(-2)$ and $f(2) = g(2)$ . This is to be expected; take a step back and think why this makes sense. If you still have questions, ask your instructor!

From our list of candidates along each part of the boundary and remembering the value the function takes at the interior critical point, we see that the absolute extrema are as follows.

The absolute maximum value $F(x,y)$ takes on $R$ is $\answer {73}$ and occurs at $(x,y) = \left ( \answer {2} , \answer {4} \right )$ .
The absolute minimum value $F(x,y)$ takes on $R$ is $\answer {-6 \sqrt {3}+1}$ and occurs at $(x,y) = \left ( \answer {0} , \answer {\sqrt {3}} \right )$ .