MATH1251 Questions HELP (1 Viewer)

HeyJes · Oct 12, 2016

I feel like this is HSC again haha just with different questions

Spirits the same

InteGrand · Oct 12, 2016

1008 said:
(a) Using the notation in the suggestion provided, I get a'(x) = 1-2x and b'(x) = 2y.

If I then set these to equal 0, I get x = 1/2 and y = 0. I also used parts of my original approach to determine the other critical points. However, none of these provide the maximum value that is provided in the answers, at the point (1/2, 2) = 17/4, while the maximum I get is at (0,2) = 4. However, my minimas match.

(b) Sorry, but I have no idea how to approach this one...

Note for a), you just need to maximise and minimise x - x^2 for x in [0, 3] and maximise and minimise y^2 for y in [0,2] (the latter obviously has min. value 0 and max. value 4, no real calculations needed). Add the max. values to get f(x,y)'s max. and add the min. values to get the min.

When you found the stationary points of a(x) and b(y), the stat. point for a(x) was a max., but for b(y) was a min. So adding these isn't helpful, since we need to add min's and add max's.

InteGrand · Oct 12, 2016

1008 said:
(a) Using the notation in the suggestion provided, I get a'(x) = 1-2x and b'(x) = 2y.

If I then set these to equal 0, I get x = 1/2 and y = 0. I also used parts of my original approach to determine the other critical points. However, none of these provide the maximum value that is provided in the answers, at the point (1/2, 2) = 17/4, while the maximum I get is at (0,2) = 4. However, my minimas match.

(b) Sorry, but I have no idea how to approach this one...

$\noindent For the second one, recall that $f(x,y) = xy \left(1-(x+y)\right)$. The domain is a triangle with $x,y \geq 0$ and due to the given vertices, $x+y$ takes values between $0$ and $2$. Consider $f(x,y)$ on the family of lines $x+y = c$ (and only within the triangle) for $c$ between $0$ and $2$ (these are lines parallel to the triangle hypothesise, starting from the origin and going to the hypotenuse). On such a line, $x+y = c$, and in the triangle, $xy \leq \left(\frac{x+y}{2}\right)^2 = \frac{c^2}{4}$, with equality iff $x=y$ (by the AM-GM Inequality).$

$\noindent So on these lines, $f = xy(1-c)$. So $f$ will be non-negative if $c<1$ and non-positive if $c>1$. So it is clear we will look for the maximum of $f$ when $c<1$ and the minimum when $c>1$. For any given $c$ with $0 \leq c \leq 1$, the $1-c$ factor is non-negative, and $xy$ is always non-negative in the triangle. The max. value of $f$ is thus $\frac{c^2}{4} (1-c)$ (occurring where $y=x$ on the line $x+y = c$, which is clearly within the triangle). So to get the max. value of $f$ for all $c$ between $0$ and $1$ (which is the max. value overall clearly, since $f$ is negative if $c>1$), we just need to maximise the function $g(c) = \frac{c^2}{4} (1-c)$ for $c\in [0,1]$, which you can easily do.$

$\noindent Now consider when $1<c \leq 2$. Here, $f$ is always negative or $0$. So as mentioned earlier, we will only be interested in finding the minimum value of $f$ here (since the max. value of $f$ is clearly positive). Since $1-c<0$, to make $f$ as negative is possible (hence minimal), we'll want the $xy$ factor to be as large as possible, so again the function to consider will be $g(c) = \frac{c^2}{4} (1-c)$ (for $c\in (1,2]$) but this time we want its \emph{minimum} value for this range of $c$, and this will give us the minimum value of $f$ in the domain.$

leehuan · Oct 15, 2016

$\text{By examining the behaviour of the function close to the critical point (0,0), classify the critical point.}$

It was easy to state that:
a) x^2+y^2 ---> local min
b) x^4-y^4 ---> saddle

$\text{c) }x^2+4xy+3y^2$

Paradoxica · Oct 15, 2016

leehuan said:
$\text{By examining the behaviour of the function close to the critical point (0,0), classify the critical point.}$

It was easy to state that:
a) x^2+y^2 ---> local min
b) x^4-y^4 ---> saddle

$\text{c) }x^2+4xy+3y^2$

(x+3y)(x+y)=z

it's an oblique oriented sheet with parabolic cross-section.

leehuan · Oct 15, 2016

Paradoxica said:
(x+3y)(x+y)=z

it's an oblique oriented sheet with parabolic cross-section.

Didn't fully understand that oblique sheet language. (We didn't really name multivariable curves.)

That being said though, for a) and b) I just limited x->0 and y->0 separately. I kinda had a feeling it wouldn't work here due to the xy but I don't know how to justify it.

Paradoxica · Oct 15, 2016

leehuan said:
Didn't fully understand that oblique sheet language. (We didn't really name multivariable curves.)

That being said though, for a) and b) I just limited x->0 and y->0 separately. I kinda had a feeling it wouldn't work here due to the xy but I don't know how to justify it.

It's a special case, I think. The function is constant along one line, but convex, so it's somewhere between a stationary point and a saddle point.

InteGrand · Oct 15, 2016

leehuan said:
$\text{By examining the behaviour of the function close to the critical point (0,0), classify the critical point.}$

It was easy to state that:
a) x^2+y^2 ---> local min
b) x^4-y^4 ---> saddle

$\text{c) }x^2+4xy+3y^2$

$\noindent For (c), calling the function $f(x,y)$, it is easy to show that $(0,0)$ is a stationary point (I think they let you assume that even). Note $f(0,0) = 0$. Along the line $y=0$, we have $f(x,0) = x^{2}$. So in any neighbourhood of $(0,0)$, there exist points of the form $\left(\varepsilon,0\right)$ at which $f$ has a positive value (greater than the value at $f(0,0)$). Along the line $x = -2y$, we have $f(-2y,y) = 4y^2 + 4\times (-2y)y + 3y^2 = -y^2$. Thus in any neighbourhood of $(0,0)$ there exist points of the form $\left(-2\epsilon, \epsilon\right)$ such that at these points, $f$ is negative. Hence in any neighbourhood of the origin, there are points where $f(x,y) < f(0,0)$ and points where $f(x,y) > f(0,0)$. Therefore, $(0,0)$ is not a local extremum, so it's a saddle point.$

leehuan · Oct 15, 2016

InteGrand said:
$\noindent For (c), calling the function $f(x,y)$, it is easy to show that $(0,0)$ is a stationary point (I think they let you assume that even). Note $f(0,0) = 0$. Along the line $y=0$, we have $f(x,0) = x^{2}$. So in any neighbourhood of $(0,0)$, there exist points of the form $\left(\varepsilon,0\right)$ at which $f$ has a positive value (greater than the value at $f(0,0)$). Along the line $x = -2y$, we have $f(-2y,y) = 4y^2 + 4\times (-2y)y + 3y^2 = -y^2$. Thus in any neighbourhood of $(0,0)$ there exist points of the form $\left(-2\epsilon, \epsilon\right)$ such that at these points, $f$ is negative. Hence in any neighbourhood of the origin, there are points where $f(x,y) < f(0,0)$ and points where $f(x,y) > f(0,0)$. Therefore, $(0,0)$ is not a local extremum, so it's a saddle point.$

Ah so basically try to find some condition between x and y so that we can prove it's a min in one plane and a max in the other?

Awesome makes sense. Do we just have to guess something like x=-2y that would work?

InteGrand · Oct 15, 2016

leehuan said:
Ah so basically try to find some condition between x and y so that we can prove it's a min in one plane and a max in the other?

Awesome makes sense. Do we just have to guess something like x=-2y that would work?

$\noindent It's easier to spot by keeping Paradoxica's factorisation in mind: since $f(x,y) = (x+3y)(x+y)$, we can easily find lines of the form $x = ky$ such that $f$ will be made positive or made negative. E.g. taking $x = -2y$, we have $f(x,y) = (-2y+3y)(-2y+y) = -y^2$. Etc.$

1008 · Oct 15, 2016

InteGrand said:
$\noindent For the second one, recall that $f(x,y) = xy \left(1-(x+y)\right)$. The domain is a triangle with $x,y \geq 0$ and due to the given vertices, $x+y$ takes values between $0$ and $2$. Consider $f(x,y)$ on the family of lines $x+y = c$ (and only within the triangle) for $c$ between $0$ and $2$ (these are lines parallel to the triangle hypothesise, starting from the origin and going to the hypotenuse). On such a line, $x+y = c$, and in the triangle, $xy \leq \left(\frac{x+y}{2}\right)^2 = \frac{c^2}{4}$, with equality iff $x=y$ (by the AM-GM Inequality).$

$\noindent So on these lines, $f = xy(1-c)$. So $f$ will be non-negative if $c<1$ and non-positive if $c>1$. So it is clear we will look for the maximum of $f$ when $c<1$ and the minimum when $c>1$. For any given $c$ with $0 \leq c \leq 1$, the $1-c$ factor is non-negative, and $xy$ is always non-negative in the triangle. The max. value of $f$ is thus $\frac{c^2}{4} (1-c)$ (occurring where $y=x$ on the line $x+y = c$, which is clearly within the triangle). So to get the max. value of $f$ for all $c$ between $0$ and $1$ (which is the max. value overall clearly, since $f$ is negative if $c>1$), we just need to maximise the function $g(c) = \frac{c^2}{4} (1-c)$ for $c\in [0,1]$, which you can easily do.$

$\noindent Now consider when $1<c \leq 2$. Here, $f$ is always negative or $0$. So as mentioned earlier, we will only be interested in finding the minimum value of $f$ here (since the max. value of $f$ is clearly positive). Since $1-c<0$, to make $f$ as negative is possible (hence minimal), we'll want the $xy$ factor to be as large as possible, so again the function to consider will be $g(c) = \frac{c^2}{4} (1-c)$ (for $c\in (1,2]$) but this time we want its \emph{minimum} value for this range of $c$, and this will give us the minimum value of $f$ in the domain.$

Thanks for clarifying!

This time from Algebra:
Parts (b) and (c) of this one please

leehuan · Oct 15, 2016

When doing these questions regarding max min for multivariable functions, would we be safe to just ignore any saddle points we come across?

1008 said:
Thanks for clarifying!

This time from Algebra:
Parts (b) and (c) of this one please

Hint for part b): Expand that thing out first before you use the roots of unity sum (GP)

InteGrand · Oct 16, 2016

1008 said:
Thanks for clarifying!

This time from Algebra:
Parts (b) and (c) of this one please

Part (c) was done here: http://community.boredofstudies.org...h1231-1241-1251-sos-thread-2.html#post7184625 .

(For part b), remember sum of roots is 0, since the z^{n-1} coefficient in the polynomial is 0.)

leehuan · Oct 16, 2016

With those questions already posted, could they be done using Lagrange multipliers in a more quicker way? Or is that a meh choice because you still have to split up the cases

Paradoxica · Oct 16, 2016

leehuan said:
With those questions already posted, could they be done using Lagrange multipliers in a more quicker way? Or is that a meh choice because you still have to split up the cases

Lagrange Multipliers I see as a last resort, as it takes a mess and a half to get the answer.

seanieg89 · Oct 16, 2016

Lagrange multipliers is how you deal with constrained optimisation problems. I.e. when you are optimising a function f on a hypersurface in R^n rather than on R^n itself.

In this case however, we were optimising a function over a domain in R^n. You can optimise the function in the interior of such domains by looking at stationary points and using higher derivative tests.

To look at what the function does on the boundary, you can either locally parametrise the boundary to reduce it to a lower dimensional interior optimisation problem of the type above, or you can use Lagrange multipliers, but this is messier. The main strength of Lagrange multipliers is that we can use it to deal with implicitly defined hypersurfaces that are not easy to parametrise.

In the problem above where we optimised over a triangle, the method I hinted at and Integrand fleshed out involved writing the triangle as a union of hypersurfaces. Indeed we could use Lagrange multipliers to optimise over the individual hypersurfaces, but the whole point of the decomposition was that the function took a very simple form on the hypersurfaces and was easy to optimise with just AM-GM. Otherwise there would be no point in decomposing the domain in this way and one would just optimise over the interior and boundary points as described above.

leehuan · Oct 16, 2016

Oh right right, my bad.

Anyway, on the note of Lagrange multipliers

$\text{Find the points on the ellipse }x^2+xy+y^2=2\text{ that are closest to the origin.}$

So from the theorem I arrived at the points
$(x,y)=(\pm \sqrt{2}, \mp \sqrt{2}), \left(\pm \sqrt{\frac23}, \pm \sqrt{\frac 23}\right)$

All I want to know is what's the quickest way to finish of the question to deduce an answer?

Do we just evaluate f(x,y) (where f(x)=x^2+y^2) and just inspect the answer from there? Or did I forget something.

InteGrand · Oct 16, 2016

leehuan said:
Oh right right, my bad.

Anyway, on the note of Lagrange multipliers

$\text{Find the points on the ellipse }x^2+xy+y^2=2\text{ that are closest to the origin.}$

So from the theorem I arrived at the points
$(x,y)=(\pm \sqrt{2}, \mp \sqrt{2}), \left(\pm \sqrt{\frac23}, \pm \sqrt{\frac 23}\right)$

All I want to know is what's the quickest way to finish of the question to deduce an answer?

Do we just evaluate f(x,y) (where f(x)=x^2+y^2) and just inspect the answer from there? Or did I forget something.

Once we have the candidate points, yeah, just evaluate the function at them to find the maximum and minimum. (In this case we're optimising a continuous function over a compact domain, so we're guaranteed a global maximum and a minimum value on this domain.)

Paradoxica · Oct 16, 2016

leehuan said:
Oh right right, my bad.

Anyway, on the note of Lagrange multipliers

$\text{Find the points on the ellipse }x^2+xy+y^2=2\text{ that are closest to the origin.}$

So from the theorem I arrived at the points
$(x,y)=(\pm \sqrt{2}, \mp \sqrt{2}), \left(\pm \sqrt{\frac23}, \pm \sqrt{\frac 23}\right)$

All I want to know is what's the quickest way to finish of the question to deduce an answer?

Do we just evaluate f(x,y) (where f(x)=x^2+y^2) and just inspect the answer from there? Or did I forget something.

I used AM-GM to arrive at xy = 2/3 in order for a minimum to occur. Only the second satisfies that condition, so it must be the minimum.

leehuan · Oct 18, 2016

Use double integration to find the area of the region bounded by y=x^3 and y=x^2

Just set up the integral please

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