Find the discriminant $D\left(3,2\right)$
of a function $f\left(x,y\right)$
if
$${f}_{xx}\left(3,2\right)=-5,\phantom{\rule{1em}{0ex}}{f}_{yy}\left(3,2\right)=-1,\phantom{\rule{1em}{0ex}}{f}_{xy}\left(3,2\right)=-2.$$
Exactly one option must be correct)
Remember that $D\left(x,y\right)={f}_{xx}\left(x,y\right)\times {f}_{yy}\left(x,y\right)-{\left({f}_{xy}\left(x,y\right)\right)}^{2}.$

*Choice (a) is incorrect*

Remember that $D\left(x,y\right)={f}_{xx}\left(x,y\right)\times {f}_{yy}\left(x,y\right)-{\left({f}_{xy}\left(x,y\right)\right)}^{2}.$

*Choice (b) is incorrect*

Remember that $D\left(x,y\right)={f}_{xx}\left(x,y\right)\times {f}_{yy}\left(x,y\right)-{\left({f}_{xy}\left(x,y\right)\right)}^{2}.$

*Choice (c) is incorrect*

Remember that $D\left(x,y\right)={f}_{xx}\left(x,y\right)\times {f}_{yy}\left(x,y\right)-{\left({f}_{xy}\left(x,y\right)\right)}^{2}.$

*Choice (d) is correct!*

We have $D\left(3,2\right)={f}_{xx}\left(3,2\right)\times {f}_{yy}\left(3,2\right)-{\left({f}_{xy}\left(3,2\right)\right)}^{2}=-5\times -1-{\left(-2\right)}^{2}=1.$

*Choice (e) is incorrect*

You are given that $\left(1,4\right)$
is a critical point of $g\left(x,y\right)$.
What type of critical point is it if the second partial derivatives have the values
below?
$${g}_{xx}\left(1,4\right)=2,\phantom{\rule{1em}{0ex}}{g}_{yy}\left(1,4\right)=4,\phantom{\rule{1em}{0ex}}{g}_{xy}\left(1,4\right)=3.$$
Exactly one option must be correct)

*Choice (a) is incorrect*

Look up the second partial
derivative test for functions of two variables in the lecture notes or in the tutorial
for week 8.

*Choice (b) is incorrect*

Look up the second partial derivative test for
functions of two variables in the lecture notes or in the tutorial for week
8.

*Choice (c) is correct!*

$D\left(1,4\right)=2\times 4-{3}^{2}=-1<0$.
When the discriminant is negative, the critical point is a saddle point.

*Choice (d) is incorrect*

Look up the second partial derivative test for
functions of two variables in the lecture notes or in the tutorial for week 8.

You are given that $\left(2,-1\right)$
is a critical point of $g\left(x,y\right)$.
What type of critical point is it if the second partial derivatives have the values
below?
$${g}_{xx}\left(2,-1\right)=-12,\phantom{\rule{1em}{0ex}}{g}_{yy}\left(2,-1\right)=-4,\phantom{\rule{1em}{0ex}}{g}_{xy}\left(2,-1\right)=5.$$
Exactly one option must be correct)
Look up the second partial
derivative test for functions of two variables in the lecture notes or in the tutorial for
week 8.
Look up the second partial derivative test for functions of two variables
in the lecture notes or in the tutorial for week 8.
Look up the second partial derivative test for
functions of two variables in the lecture notes or in the tutorial for week
8.

*Choice (a) is correct!*

$D\left(2,-1\right)=-12\times -4-{5}^{2}=23>0$. When the discriminant
is positive and ${f}_{xx}<0$,
the critical point is a local maximum.

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is incorrect*

Suppose $f\left(x,y\right)$
is a function of two variables. If the critical point
$\left(a,b\right)$ is a local
maximum point, which of the following are true statements? (Zero or more options can
be correct)

For example, choice (a) should be True.

For example, choice (b) should be False.

For example, choice (c) should be False.

For example, choice (d) should be False.

For example, choice (e) should be True.

*There is at least one mistake.*

For example, choice (a) should be True.

This is true, and in fact ${f}_{y}\left(a,b\right)$
must be zero as well.

*There is at least one mistake.*

For example, choice (b) should be False.

*There is at least one mistake.*

For example, choice (c) should be False.

*There is at least one mistake.*

For example, choice (d) should be False.

*There is at least one mistake.*

For example, choice (e) should be True.

This is true,and in fact, they must both be negative.

*Correct!*

*True*This is true, and in fact ${f}_{y}\left(a,b\right)$ must be zero as well.*False**False**False**True*This is true,and in fact, they must both be negative.

The function $f$ of two
variables given by $f\left(x,y\right)=3{x}^{2}+4xy+4{y}^{2}+2x+12y+5$
has Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is correct!*

The partial derivatives ${f}_{x}$
and ${f}_{y}$
are ${f}_{x}=6x+4y+2$
and ${f}_{y}=4x+8y+12.$
They are both equal to zero at a critical point. Solving
${f}_{x}=0$ and
${f}_{y}=0$ simultaneously gives
the single critical point $\left(1,-2\right).$

*Choice (e) is incorrect*

Find the lowest point on the surface given by
$z=3{x}^{4}+4{x}^{3}+6{y}^{4}-16{y}^{3}+12{y}^{2}+7.$ Type
the $z$
value of this point into the box.

*Correct!*

We have $\frac{\partial z}{\partial x}=12{x}^{3}+12{x}^{2}$
and $\frac{\partial z}{\partial y}=24{y}^{3}-48{y}^{2}+24y.$
Setting each of these equal to zero and solving gives four critical points
$\left(0,0\right),\phantom{\rule{0.3em}{0ex}}\left(-1,0\right),\phantom{\rule{0.3em}{0ex}}\left(0,1\right),\phantom{\rule{0.3em}{0ex}}\left(-1,1\right).$ The smallest of the
function values ($z$ values)
at each of these points is $6$,
at the point $\left(-1,0\right)$.
The second derivative test shows that this is a local minimum.
The terms of highest degree in the function formula are
$3{x}^{4}$ and
$6{y}^{4}$, which become extremely
large and positive as $x$
and $y$
become large in any direction. This tells us that the local minimum at
$\left(-1,0\right)$ is
also the global minimum.

*Incorrect.*

*Please try again.*

Check that your partial derivatives are correct. You should have
$\frac{\partial z}{\partial x}=12{x}^{3}+12{x}^{2}$ and
$\frac{\partial z}{\partial y}=24{y}^{3}-48{y}^{2}+24y.$ Setting
each of these equal to zero and solving gives four critical points. Find the function values
($z$
values) at each of these and select the smallest for your answer. (Why?)

Find the highest point on the surface given by
$z=\frac{1}{1-x+y+{x}^{2}+{y}^{2}}.$ Type
the $z$
value of this point into the box.

*Correct!*

We have
$$\frac{\partial z}{\partial x}=-\frac{-1+2x}{{\left(1-x+y+{x}^{2}+{y}^{2}\right)}^{2}}$$

and

$$\frac{\partial z}{\partial y}=-\frac{1+2y}{{\left(1-x+y+{x}^{2}+{y}^{2}\right)}^{2}}.$$Hence $\left(\frac{1}{2},-\frac{1}{2}\right)$ is the only critical point and here the function value is $2$. The second derivative test shows that this is a local maximum. This must also be the global maximum. (Why?)

*Incorrect.*

*Please try again.*

Check that your partial derivatives are correct. You should have
$\frac{\partial z}{\partial x}=-\frac{-1+2x}{{\left(1-x+y+{x}^{2}+{y}^{2}\right)}^{2}}$ and
$\frac{\partial z}{\partial y}=-\frac{1+2y}{{\left(1-x+y+{x}^{2}+{y}^{2}\right)}^{2}}.$ Setting
each of these equal to zero and solving gives one critical point. Find the function
value ($z$
value) at this point and this will be the required answer. (Why?)

Find $\left(a,b\right)$
so that the corresponding point of the surface given by
$x+y+z=3$
is closest to the origin. Exactly one option must be correct)
We wish to minimise the square of the distance from the origin to the point
$\left(x,y,z\right)$ on the surface; that is,
we wish to minimise ${x}^{2}+{y}^{2}+{z}^{2}$,
where $z=3-x-y$.
The function to be minimised is therefore
$$F\left(x,y\right)={x}^{2}+{y}^{2}+{\left(3-x-y\right)}^{2}=2{x}^{2}+2{y}^{2}-6x-6y+2xy+9.$$
Now find the critical points of $F$.

*Choice (a) is incorrect*

We
wish to minimise the square of the distance from the origin to the point
$\left(x,y,z\right)$ on the surface; that is,
we wish to minimise ${x}^{2}+{y}^{2}+{z}^{2}$,
where $z=3-x-y$.
The function to be minimised is therefore
$$F\left(x,y\right)={x}^{2}+{y}^{2}+{\left(3-x-y\right)}^{2}=2{x}^{2}+2{y}^{2}-6x-6y+2xy+9.$$
Now find the critical points of $F$.

*Choice (b) is incorrect*

We
wish to minimise the square of the distance from the origin to the point
$\left(x,y,z\right)$ on the surface; that is,
we wish to minimise ${x}^{2}+{y}^{2}+{z}^{2}$,
where $z=3-x-y$.
The function to be minimised is therefore
$$F\left(x,y\right)={x}^{2}+{y}^{2}+{\left(3-x-y\right)}^{2}=2{x}^{2}+2{y}^{2}-6x-6y+2xy+9.$$
Now find the critical points of $F$.

*Choice (c) is incorrect*

We
wish to minimise the square of the distance from the origin to the point
$\left(x,y,z\right)$ on the surface; that is,
we wish to minimise ${x}^{2}+{y}^{2}+{z}^{2}$,
where $z=3-x-y$.
The function to be minimised is therefore
$$F\left(x,y\right)={x}^{2}+{y}^{2}+{\left(3-x-y\right)}^{2}=2{x}^{2}+2{y}^{2}-6x-6y+2xy+9.$$
Now find the critical points of $F$.

*Choice (d) is correct!*

We wish to minimise ${x}^{2}+{y}^{2}+{z}^{2}$,
where $z=3-x-y$.
The function to be minimised is therefore
$$F\left(x,y\right)=2{x}^{2}+2{y}^{2}-6x-6y+2xy+9.$$
The only critical point is $\left(1,1\right)$
and this is a local minimum by the second derivative test.

*Choice (e) is incorrect*

The function given by $f\left(x,y\right)={x}^{4}+{y}^{4}-4xy+1$
has Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is correct!*

The partial derivatives ${f}_{x}$ and
${f}_{y}$ are simultaneously zero
at three critical points $\left(0,0\right),\phantom{\rule{0.3em}{0ex}}\left(1,1\right),\phantom{\rule{0.3em}{0ex}}\left(-1,-1\right)$.
The second derivative test shows that there is a saddle point at the origin and local
minima at the other two points.

*Choice (e) is incorrect*

What is the absolute maximum value of
$f\left(x,y\right)=1+4x-5y$ on
the domain consisting of the interior and boundary of the triangle with vertices
$\left(0,0\right),\phantom{\rule{0.3em}{0ex}}\left(2,0\right),\phantom{\rule{0.3em}{0ex}}\left(0,3\right)$ ? Exactly one option
must be correct)
Find
any critical points of the function which lie inside the domain, and evaluate the
function values there. Then check the function values on the boundary of the
domain.

*Choice (a) is incorrect*

Find any critical points of the function which lie inside the domain, and evaluate the
function values there. Then check the function values on the boundary of the domain.

*Choice (b) is incorrect*

Find
any critical points of the function which lie inside the domain, and evaluate the
function values there. Then check the function values on the boundary of the
domain.

*Choice (c) is incorrect*

Find
any critical points of the function which lie inside the domain, and evaluate the
function values there. Then check the function values on the boundary of the
domain.

*Choice (d) is incorrect*

*Choice (e) is correct!*

In fact, the maximum value is
$9$ and it occurs
at the vertex $\left(2,0\right)$.
There are no critical points inside the domain so we must find the maximum
values on the boundary of the domain. On the vertical boundary, where
$x=0$, the function
simplifies to $f\left(0,y\right)=1-5y$,
for values of $y$ in
the interval $\left[0,3\right]$.
The maximum value on this boundary occurs at
$\left(x,y\right)=\left(0,0\right)$ and gives a function value
of $1$. On the horizontal
boundary, where $y=0$, the
function simplifies to $f\left(x,0\right)=1+4x$,
for values of $x$ in
the interval $\left[0,2\right]$.
The maximum value on this boundary occurs at
$\left(x,y\right)=\left(2,0\right)$ and gives a
function value of $9$.
On the other boundary (corresponding to the line
$y=-\frac{3}{2}x+3$) the
function simplifies to
$$f\left(x,y\right)=1+4x-5\left(-\frac{3}{2}x+3\right)=\frac{23}{2}\phantom{\rule{0.3em}{0ex}}x-14$$
for values of $x$ in the
interval $\left[0,2\right]$. The maximum
value occurs at $\left(2,0\right)$
and is $23-14=9$.
Therefore the absolute maximum function value on the given domain is
$9$.