## MATH1001 Quizzes

Quiz 7: Optimizing functions of two variables
Question 1 Questions
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)
 a) $-9$ b) $9$ c) $-1$ d) $1$ e) None of the above.

Choice (a) is incorrect
Remember that $D\left(x,y\right)={f}_{xx}\left(x,y\right)×{f}_{yy}\left(x,y\right)-{\left({f}_{xy}\left(x,y\right)\right)}^{2}.$
Choice (b) is incorrect
Remember that $D\left(x,y\right)={f}_{xx}\left(x,y\right)×{f}_{yy}\left(x,y\right)-{\left({f}_{xy}\left(x,y\right)\right)}^{2}.$
Choice (c) is incorrect
Remember that $D\left(x,y\right)={f}_{xx}\left(x,y\right)×{f}_{yy}\left(x,y\right)-{\left({f}_{xy}\left(x,y\right)\right)}^{2}.$
Choice (d) is correct!
We have $D\left(3,2\right)={f}_{xx}\left(3,2\right)×{f}_{yy}\left(3,2\right)-{\left({f}_{xy}\left(3,2\right)\right)}^{2}=-5×-1-{\left(-2\right)}^{2}=1.$
Choice (e) is incorrect
Remember that $D\left(x,y\right)={f}_{xx}\left(x,y\right)×{f}_{yy}\left(x,y\right)-{\left({f}_{xy}\left(x,y\right)\right)}^{2}.$
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)
 a) Local maximum. b) Local minimum. c) Saddle point. d) Not enough information to decide.

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!
$D\left(1,4\right)=2×4-{3}^{2}=-1<0$. 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)
 a) Local maximum. b) Local minimum. c) Saddle point. d) Not enough information to decide.

Choice (a) is correct!
$D\left(2,-1\right)=-12×-4-{5}^{2}=23>0$. When the discriminant is positive and ${f}_{xx}<0$, the critical point is a local maximum.
Choice (b) is incorrect
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 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 (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.
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)
 a) ${f}_{x}\left(a,b\right)$ must be zero. b) ${f}_{x}\left(a,b\right)$ and ${f}_{y}\left(a,b\right)$ must be both positive or both negative. c) ${f}_{xx}\left(a,b\right)$ and ${f}_{yy}\left(a,b\right)$ must both be positive. d) ${f}_{xx}\left(a,b\right)$ must be negative and ${f}_{yy}\left(a,b\right)$ must be positive. e) ${f}_{xx}\left(a,b\right)$ and ${f}_{yy}\left(a,b\right)$ must have the same sign.

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!
1. True This is true, and in fact ${f}_{y}\left(a,b\right)$ must be zero as well.
2. False
3. False
4. False
5. 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)
 a) one critical point at $\left(0,6\right)$ b) two critical points, at $\left(1,-1\right)$ and $\left(2,3\right)$ c) two critical points, at $\left(1,2\right)$ and $\left(3,1\right)$ d) one critical point at $\left(1,-2\right)$ e) none of the above.

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.
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?)

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)
 a) $\left(-1,0\right)$ b) $\left(2,1\right)$ c) $\left(1,0\right)$ d) $\left(1,1\right)$ e) None of the above

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
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 $F$.
The function given by $f\left(x,y\right)={x}^{4}+{y}^{4}-4xy+1$ has Exactly one option must be correct)
 a) two saddle points b) one local minimum and two saddle points c) one local maximum and one saddle point d) two local minima and one saddle point e) None of the above

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)
 a) $5$ b) $11$ c) $3$ d) $6$ e) None of the above

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
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 (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 $9$.