MHB Multivariable optimization with constraint

[Petrus]

Calculate biggest and lowest value to function
$$f(x,y)=x^5y^4e^{-3x-3y}$$
In the triangle has vertices in points $$\left(0,0 \right)$$,$$\left(6,0 \right)$$ and $$\left(0,6 \right)$$
Before I start I want to warn that I used google translate in the text 'In the triangle has vertices in points'
Progress:
I start to derivate (I need confirm if I did right when I did it)
Derivate x: $$5x^4y^4e^{-3x-3y}(-3-3y)$$
Derivate y: $$x^5 4y^3e^{-3x-3y}(-3x-3)$$

 

[alyafey22]

Re: Two variable, e

I think you are trying to find the derivative with respect to x and the derivative with respect to y , right ?

To do so you must treat the other variable as constant ...

 

[Petrus]

Re: Two variable, e

I think you are trying to find the derivative with respect to x and the derivative with respect to y , right ?

To do so you must treat the other variable as constant ...

I think I got it now, I forgot to use product and chain rule..
Respect to x: $$5x^4y^4e^{-3x-3y}-x^5y^4e^-3{-3x-3y}$$
Respect to y: $$x^54y^3e^{-3x-3y}-x^5y^4e^-3{-3x-3y}$$
I am correct?

 

[alyafey22]

Re: Two variable, e

I think I got it now, I forgot to use product and chain rule..
Respect to x: $$5x^4y^4e^{-3x-3y}-x^5y^4e^{-3x-3y}-3$$
Respect to y: $$x^54y^3e^{-3x-3y}-x^5y^4e^{-3x-3y}-3$$
I am correct?

How did you get -3 at the end ?

 

[Petrus]

Re: Two variable, e

How did you get -3 at the end ?

Sorry forgot to write... I did edit my post. Next I shall find critical point and that I got problem... I know that I have to rewrite y as a function of x..

 

[alyafey22]

Re: Two variable, e

what is the derivative of $$e^{-3x-3y}$$ with respect to x ?

 

[Petrus]

Re: Two variable, e

what is the derivative of $$e^{-3x-3y}$$ with respect to x ?

$$e^{-3x-3y}(-3-3y)$$

 

[Petrus]

Re: Two variable, e

$$e^{-3x-3y}(-3-3y)$$

That's wrong... It should be $$-3e^{-3x-3y}$$

 

[Petrus]

Respect to x: $$5x^4y^4e^{-3x-3y}+x^5y^43e^{-3x-3y}$$
Respect to y: $$x^54y^3e^{-3x-3y}+x^5y^43e^{-3x-3y}$$
And now I have to find critical point.
If we start with respect to y
I can factor out $$x^5e^{3x-3y}$$
so I got $$x^5e^{3x-3y}(4y^3+3y^4)$$ I am correct?

 

[MarkFL]

You are given the function:

$$f(x,y)=x^5y^4e^{-3(x+y)}$$

To find the critical values, you want to equate the partial derivatives to zero:

i) $$f_x(x,y)=y^4\left(x^5\left(e^{-3(x+y)}(-3) \right)+5x^4e^{-3(x+y)} \right)=x^4y^4e^{-3(x+y)}(5-3x)=0$$

ii) $$f_y(x,y)=x^5\left(y^4\left(e^{-3(x+y)}(-3) \right)+4y^3e^{-3(x+y)} \right)=x^5y^3e^{-3(x+y)}(4-3y)=0$$

What are your critical or stationary points? For each point within the given triangle, you may now use the second partials test for relative extrema to determine that nature of the extrema. Lastly, you want to check the boundary extrema:

Recall that if a function $f$ of a single variable $x$ is continuous on a closed interval $[a,b]$, then $f$ always has absolute extrema on the interval. In this case, an absolute extremum could be an endpoint extremum. For a function $f$ of two variables $x$ and $y$, the Extreme Value Theorem states that if $f$ is continuous on a closed and bounded region $R$, then $f$ has absolute extrema at points in $R$. Analogous to endpoint extrema, a function of two variables could have boundary extrema.

Can you give the boundaries of the region $R$ in this case?

 

[Petrus]

You are given the function:

$$f(x,y)=x^5y^4e^{-3(x+y)}$$

To find the critical values, you want to equate the partial derivatives to zero:

i) $$f_x(x,y)=y^4\left(x^5\left(e^{-3(x+y)}(-3) \right)+5x^4e^{-3(x+y)} \right)=x^4y^4e^{-3(x+y)}(5-3x)=0$$

ii) $$f_y(x,y)=x^5\left(y^4\left(e^{-3(x+y)}(-3) \right)+4y^3e^{-3(x+y)} \right)=x^5y^3e^{-3(x+y)}(4-3y)=0$$

What are your critical or stationary points? For each point within the given triangle, you may now use the second partials test for relative extrema to determine that nature of the extrema. Lastly, you want to check the boundary extrema:

Recall that if a function $f$ of a single variable $x$ is continuous on a closed interval $[a,b]$, then $f$ always has absolute extrema on the interval. In this case, an absolute extremum could be an endpoint extremum. For a function $f$ of two variables $x$ and $y$, the Extreme Value Theorem states that if $f$ is continuous on a closed and bounded region $R$, then $f$ has absolute extrema at points in $R$. Analogous to endpoint extrema, a function of two variables could have boundary extrema.

Can you give the boundaries of the region $R$ in this case?

Is this correct?
Open: $$0<x<6$$, $$0<y<6$$
Closed: $$0≤x≤6$$, $$0≤y≤6$$
Our tangent line is $$y=-x+6$$

 

[Jameson]

Your boundaries are correct. You need to find the critical values of $f(x,y)$ using what MarkFL wrote in his last post, which is to solve $$x^4y^4e^{-3(x+y)}(5-3x)=0$$ and $$x^5y^3e^{-3(x+y)}(4-3y)=0$$ for all $(x,y)$ pairs in the region of the triangle. Then you need to determine the nature of the critical points. Have you done this before? Here is a useful webpage that goes over all of this.

 

[Petrus]

Your boundaries are correct. You need to find the critical values of $f(x,y)$ using what MarkFL wrote in his last post, which is to solve $$x^4y^4e^{-3(x+y)}(5-3x)=0$$ and $$x^5y^3e^{-3(x+y)}(4-3y)=0$$ for all $(x,y)$ pairs in the region of the triangle. Then you need to determine the nature of the critical points. Have you done this before? Here is a useful webpage that goes over all of this.

Hello Jameson!
Don't get me wrong but I do understand the progress I am just having big problem even thought Mark gave me big respons about critical point but I still got problem with it... I only done simpel problem, I understand that I Will find a subsitate for exempel y and put in the other equation. I am having problem with this one.

edit: I Will post some calcululate without Max,min value

 

[Jameson]

Hi Petrus. We can simplify the two equations Mark gave you. :)

$$x^4y^4e^{-3(x+y)}(5-3x)=0$$ can be simplified to $$(5-3x)=0$$ in this problem. Why? Since $x,y >0$ that means $x^4 y^4 e^{-3(x+y)}$ cannot be 0 in that domain, so we can cancel out all of those things and we're left with something much simpler. You can do the same thing for the second equation. Once you do that, you should solve the two equations for the critical points.

 

[Petrus]

$$$$

Hi Petrus. We can simplify the two equations Mark gave you. :)

$$x^4y^4e^{-3(x+y)}(5-3x)=0$$ can be simplified to $$(5-3x)=0$$ in this problem. Why? Since $x,y >0$ that means $x^4 y^4 e^{-3(x+y)}$ cannot be 0 in that domain, so we can cancel out all of those things and we're left with something much simpler. You can do the same thing for the second equation. Once you do that, you should solve the two equations for the critical points.

Hello Jameson,
I did understand but I did not understand what you mean with $$x,y>0$$ what domain did you talk about? e^0=1 so that works?

 

[Jameson]

$$$$
Hello Jameson,
I did understand but I did not understand what you mean with $$x,y>0$$ what domain did you talk about? e^0=1 so that works?

I want to say I am not 100% sure about this since it's been some time since I took this course, but I'll explain my thoughts.

The triangle has three boundary lines and is a closed region. We will test the boundaries separately though at the end of the problem because we can't use calculus to solve for that like we can for the relative maximums and minimums. So we can just focus on the interior of the triangle for now which means that $x>0$ and $y>0$.

Anyway, so if $x,y>0$ that means neither one can be 0 obviously. Now if we look at $$x^4y^4e^{-3(x+y)}(5-3x)=0$$ we notice that there are two groups of terms. One group is $$x^4y^4e^{-3(x+y)}$$ and the other is $$(5-3x)$$. Why? All of these things are multiplied together so whenever anyone term equals 0, the whole thing equals 0. As we said though, $x \ne 0$, $y \ne 0$ and $e^{x}$ is never negative, so these terms can't help us find when the expression is 0 and we can remove them.

Only the last terms in both equations can be used to solve which simplifies the math quite a lot. What is left is $5-3x=0$ and $4-3y=0$. So from those two equations we should have a critical point to check now.

 

[Jameson]

I think I was wrong actually about not including 0. :(

You can include $x=0$, $y=0$ in the region. If you do then you get another point for the derivative equal to zero, the point (0,0). So you should have two points to test from the derivative calculations, $(0,0)$ and $$\left( \frac{5}{3}, \frac{4}{3} \right)$$.

Now you need to test the boundaries of the triangle. I know this is the correct method for doing this but in this problem two equations are always 0 and I need to check how to interpret that but here's the math. We start with $$f(x,y)=x^5y^4e^{-3x-3y}$$. The triangle has one leg on the x-axis from [0,6] so for that leg, $y=0$. We can plug that into $f(x,y)$ to define a new function for this leg. $$f(x,0)=0*y^4e^{-3x-3y}=0$$ and it seems the same thing happens for $f(0,y)$ when we test the other leg.

So all that's left is the third leg, which is $y=-x+6$ as you correctly stated. I think you'll have to use Lagrange multipliers for this one since there are two variables. I don't think this post has any new information for you but I just wanted to make sure.

So what's left is to find the critical points of $x+y=6$ over the region of the triangle and test the other points we've found. Hopefully MarkFL will be on soon to steer the thread back on course. Sorry I'm a little out of practice with this topic but hopefully something has been useful. :)

 

[Petrus]

I think I was wrong actually about not including 0. :(

You can include $x=0$, $y=0$ in the region. If you do then you get another point for the derivative equal to zero, the point (0,0). So you should have two points to test from the derivative calculations, $(0,0)$ and $$\left( \frac{5}{3}, \frac{4}{3} \right)$$.

Now you need to test the boundaries of the triangle. I know this is the correct method for doing this but in this problem two equations are always 0 and I need to check how to interpret that but here's the math. We start with $$f(x,y)=x^5y^4e^{-3x-3y}$$. The triangle has one leg on the x-axis from [0,6] so for that leg, $y=0$. We can plug that into $f(x,y)$ to define a new function for this leg. $$f(x,0)=0*y^4e^{-3x-3y}=0$$ and it seems the same thing happens for $f(0,y)$ when we test the other leg.

So all that's left is the third leg, which is $y=-x+6$ as you correctly stated. I think you'll have to use Lagrange multipliers for this one since there are two variables. I don't think this post has any new information for you but I just wanted to make sure.

So what's left is to find the critical points of $x+y=6$ over the region of the triangle and test the other points we've found. Hopefully MarkFL will be on soon to steer the thread back on course. Sorry I'm a little out of practice with this topic but hopefully something has been useful. :)

Hello Jameson,
Cleaver cleaver, math is such so intressing how you think (There is always trick that is so logic but you never think about!) .. basicly when you solve the equation you think 2 case.
case 1 $$x,y=0$$
case 2 $$x,y>0$$
I want to add that I solved the problem min =0 and max =0.005. I want to check Mark progress on that simplify that I always get stuck on when I try:( I will try more and try understand!
Thanks to you all who respond and helped me! (Dance)

 

[MarkFL]

...

You can include $x=0$, $y=0$ in the region. If you do then you get another point for the derivative equal to zero, the point (0,0). So you should have two points to test from the derivative calculations, $(0,0)$ and $$\left( \frac{5}{3}, \frac{4}{3} \right)$$.

Yes, I agree with those critical points since they are on the region $R$, and I was wrong about needing to use the second partials test, in this case we merely need to evaluate the function at these points and take these values as possible absolute extrema.

...

Now you need to test the boundaries of the triangle. I know this is the correct method for doing this but in this problem two equations are always 0 and I need to check how to interpret that but here's the math. We start with $$f(x,y)=x^5y^4e^{-3x-3y}$$. The triangle has one leg on the x-axis from [0,6] so for that leg, $y=0$. We can plug that into $f(x,y)$ to define a new function for this leg. $$f(x,0)=0*y^4e^{-3x-3y}=0$$ and it seems the same thing happens for $f(0,y)$ when we test the other leg.

Yes, I agree, we find that the function is zero on all points of the two legs of the bounding triangle. For all other points on the triangle we will find $0<f(x,y)$, so we may conclude at this point that the absolute minimum on the given region is $f(x,y)=0$.

...

So all that's left is the third leg, which is $y=-x+6$ as you correctly stated. I think you'll have to use Lagrange multipliers for this one since there are two variables. I don't think this post has any new information for you but I just wanted to make sure.

We could use Lagrange multipliers, or we may substitute for one of the variables using the constraint to get a function in one variable and optimize by equating the derivative of that function in one variable to zero.

I think it is much simpler in this case to use Lagrange multipliers, as this simply implies equating the partials we have already computed. From this we may get an equation relating $x$ and $y$ that we may use in the constraint to get a quadratic with two real roots, giving us two possible points on the hypotenuse of the triangle as possible absolute extrema, however, care must be taken to ensure these points are on the hypotenuse (one is not hint hint). :D

 

[Jameson]

Yes, I agree with those critical points since they are on the region $R$, and I was wrong about needing to use the second partials test, in this case we merely need to evaluate the function at these points and take these values as possible absolute extrema.

I completely forgot to mention testing for a saddle point. Your point though, I think, is that since we have all of the critical points over the region then we can test all of them to find the absolute max and min over the region. If the point we found in the interior of the triangle is in fact a saddle point it wouldn't matter because it's value would fall between the max and min that we will find. Is that your reasoning?

This is a good problem for me to brush up on my Calc III knowledge. I won't need to use a lot of the material usually covered but being able to analyze functions of more than one variable is a useful skill to have.

 

[Petrus]

You are given the function:

$$f(x,y)=x^5y^4e^{-3(x+y)}$$

To find the critical values, you want to equate the partial derivatives to zero:

i) $$f_x(x,y)=y^4\left(x^5\left(e^{-3(x+y)}(-3) \right)+5x^4e^{-3(x+y)} \right)=x^4y^4e^{-3(x+y)}(5-3x)=0$$

ii) $$f_y(x,y)=x^5\left(y^4\left(e^{-3(x+y)}(-3) \right)+4y^3e^{-3(x+y)} \right)=x^5y^3e^{-3(x+y)}(4-3y)=0$$

Hello Mark,
I don't understand the last part of i) $$f_x(x,y)=y^4\left(x^5\left(e^{-3(x+y)}(-3) \right)+5x^4e^{-3(x+y)} \right)=x^4y^4e^{-3(x+y)}(5-3x)=0$$
How do you get it to $$x^4y^4e^{-3(x+y)}(5-3x)=0$$

 

[MarkFL]

I completely forgot to mention testing for a saddle point. Your point though, I think, is that since we have all of the critical points over the region then we can test all of them to find the absolute max and min over the region. If the point we found in the interior of the triangle is in fact a saddle point it wouldn't matter because it's value would fall between the max and min that we will find. Is that your reasoning?...

Yes exactly, in much the same way for a function in one variable for which a critical point is not an extremum as the second derivative is zero there. Rather than compute the second derivative to then find we may discard this point, it is simpler just to evaluate the function at that point, and then find it lies in between the actual absolute extrema at the endpoints, and discard it then.

It doesn't hurt to go ahead and find the nature of the extrema, but is simply unnecessary for problems like this. It alleviates the need to compute all of the second partials, etc.

 

[MarkFL]

Hello Mark,
I don't understand the last part of i) $$f_x(x,y)=y^4\left(x^5\left(e^{-3(x+y)}(-3) \right)+5x^4e^{-3(x+y)} \right)=x^4y^4e^{-3(x+y)}(5-3x)=0$$
How do you get it to $$x^4y^4e^{-3(x+y)}(5-3x)=0$$

Okay, let's look at:

$$f_x(x,y)=y^4\left(x^5\left(e^{-3(x+y)}(-3) \right)+5x^4e^{-3(x+y)} \right)$$

Look at the two terms within the outer parentheses, and you should notice that both have $$x^4e^{-3(x+y)}$$ as factors, so pull that factor out front to get:

$$f_x(x,y)=x^4y^4e^{-3(x+y)}\left(x(-3)+5 \right)=x^4y^4e^{-3(x+y)}(5-3x)$$

Does this make sense?

 

[Petrus]

I completely forgot to mention testing for a saddle point. Your point though, I think, is that since we have all of the critical points over the region then we can test all of them to find the absolute max and min over the region. If the point we found in the interior of the triangle is in fact a saddle point it wouldn't matter because it's value would fall between the max and min that we will find. Is that your reasoning?

This is a good problem for me to brush up on my Calc III knowledge. I won't need to use a lot of the material usually covered but being able to analyze functions of more than one variable is a useful skill to have.

Hello Jameson,
I think you are confusing yourself now:P ( I do it as well with this problem)
I don't calculate local min and max values ( that means I don't use 'Second Derivate Test')
edit: I am wrong?

 

[MarkFL]

Hello Jameson,
I think you are confusing yourself now:P ( I do it as well with this problem)
I don't calculate local min and max values ( that means I don't use 'Second Derivate Test')

No, Jameson is correctly clarifying a point I made earlier.

You DO want to find all potential local or relative extrema within the triangle as critical values to test. You also want to test the boundaries. All we were saying is that the second partials test is unnecessary.

What have you found so far just so we know where you are at and where you need to go. :D

 

[Petrus]

No, Jameson is correctly clarifying a point I made earlier.

You DO want to find all potential local or relative extrema within the triangle as critical values to test. You also want to test the boundaries. All we were saying is that the second partials test is unnecessary.

What have you found so far just so we know where you are at and where you need to go. :D

Hello Mark,
Ok I need explain. Cause the method I use was I will use citat
"To find the absolute maximum and minimum values of a continuous function f on a closed, bounded set D:
1. Find the values of f at the critical points of f in D.
2. Find the extreme values of f on the boundary of D.
3. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value."
(I want to add that I now got how you did that simplify! thanks again Mark

 

[Jameson]

That looks good, Petrus. The only thing that hasn't been done in this thread is to find any critical points on the boundary line $x+y=6$. MarkFL suggested using Lagrange multipliers to find the critical points, which I agree with. Once you find the last point (MarkFL gave us a hint that there is just one more) then you can finally plug in all of your test points and be done. One more thing to solve for and then it's just plugging in numbers for the final answer.

 

[MarkFL]

Okay, sounds like you have a good plan of attack. (Cool)

1.) Have you determined the value of $f(x,y)$ at your two critical points? If so what are they?

2.) You know that $f(x,y)=0$ on the two legs of the bounding triangle. Have you determined the critical point on the hypotenuse? Can you state the relationship between $x$ and $y$ as found by the method of Lagrange multipliers? If so, can you state a quadratic by using this relationship in the constraint given by the hypotenuse? If so can you state the roots, and show why one must be discarded? Once you do this, then evaluate the function at this boundary point. You should now have 3 values for part 3.).

3.) Once you have successfully completed 1.) and 2.) then this should be easy.

 

[Petrus]

Ok here we go (sorry for bad use of latex)
1. Our critical points are $$(0,0)$$ and $$\left(\frac{5}{3},\frac{4}{3} \right)$$

Well our critical point $$(0,0)$$ will give us the result 0. And the critical point $$\left(\frac{5}{3},\frac{4}{3} \right)$$ give us 0.005

2. As Jameson said:

$$f(0,y)=0 $$ on $$[0,6]$$
$$f(x,0)=0$$ on $$[0,6]$$

(Now for the next hardest part :()

$$f(x,6-x)=x^5*(6-x)^4e^{-3x-3(6-x)}$$ on $$[0,6]$$ (I need to find the critical points as well...)

let's call it $g(x)$
$g'(x)=?$ I am struggling with derivative that product rule when I got 3 function :S

 

[MarkFL]

Hey Petrus,

I dressed up your $\LaTeX$ just a bit. For the most part it was fine. I will address your points as follows:

...
1. Our critical points are $$(0,0)$$ and $$\left(\frac{5}{3},\frac{4}{3} \right)$$

Well our critical point $$(0,0)$$ will give us the result 0. And the critical point $$\left(\frac{5}{3},\frac{4}{3} \right)$$ give us 0.005

Yes, that is correct, although I would keep the exact value for the purpose of answering the question, and use the decimal approximation for the purpose of comparison. So, we have:

$$f(0,0)=0$$

$$f\left(\frac{5}{3},\frac{4}{3} \right)=\frac{800000}{19683e^9}\approx0.005015894084709833$$

...

2. As Jameson said:

$$f(0,y)=0 $$ on $$[0,6]$$
$$f(x,0)=0$$ on $$[0,6]$$

Yes, the function is zero at every point on the two legs. So we know the absolute minimum is zero, do you see why?

...

(Now for the next hardest part :()

$$f(x,6-x)=x^5*(6-x)^4e^{-3x-3(6-x)}$$ on $$[0,6]$$ (I need to find the critical points as well...)

let's call it $g(x)$
$g'(x)=?$ I am struggling with derivative that product rule when I got 3 function :S

This part will be made much easier if you use Lagrange multipliers. Can you show us why this would imply:

$$f_x(x,y)=f_y(x,y)$$ ?

 

[Petrus]

Hey Petrus,

I dressed up your $\LaTeX$ just a bit. For the most part it was fine. I will address your points as follows:
Yes, that is correct, although I would keep the exact value for the purpose of answering the question, and use the decimal approximation for the purpose of comparison. So, we have:

$$f(0,0)=0$$

$$f\left(\frac{5}{3},\frac{4}{3} \right)=\frac{800000}{19683e^9}\approx0.005015894084709833$$
Yes, the function is zero at every point on the two legs. So we know the absolute minimum is zero, do you see why?
This part will be made much easier if you use Lagrange multipliers. Can you show us why this would imply:

$$f_x(x,y)=f_y(x,y)$$ ?

Can you help me with start with lagrange multipliers with this function?I can't see where I shall put the landa

 

[MarkFL]

Can you help me with start with lagrange multipliers with this function?I can't see where I shall put the landa

I will show you the way I was taught to set it up.

We have the objective function $f(x,y)$ subject to the constraint $g(x,y)=x+y-6=0$

So, the method of Lagrange multipliers tells us to set up the system:

$$f_x(x,y)=\lambda\cdot g_x(x,y)$$

$$f_y(x,y)=\lambda\cdot g_y(x,y)$$

Now, what are $$g_x(x,y)$$ and $$g_y(x,y)$$ ?

 

[Petrus]

I will show you the way I was taught to set it up.

We have the objective function $f(x,y)$ subject to the constraint $g(x,y)=x+y-6=0$

So, the method of Lagrange multipliers tells us to set up the system:

$$f_x(x,y)=\lambda\cdot g_x(x,y)$$

$$f_y(x,y)=\lambda\cdot g_y(x,y)$$

Now, what are $$g_x(x,y)$$ and $$g_y(x,y)$$ ?

We got our tangent line $$y=6-x<=>6-x-y=0$$ and that is equalt to $$x+y-6=0$$
Answer to your question
$$g_x(x,y)=-5$$
$$g_y(x,y)=-5$$

 

[MarkFL]

We got our tangent line $$y=6-x<=>0=6-x-y=0$$ and that is equalt to$$x+y-6$$
Answer to your question
$$g_x(x,y)=-5$$
$$g_y(x,y)=-5$$

No, you want to take the partial derivatives of $g(x,y)$ with respect to the two variables. What you should find is:

$$g_x(x,y)=g_y(x,y)=1$$

Do you see why?

Note: I must leave for about 4.5 hours now, so if anyone else wishes to jump in that is fine with me. Petrus has informed me that he has not actually studied Lagrange multipliers yet, but he looked it up and wants to use it here. I have also told him once this is completed, I would help him also finish the problem by using the constraint to express $f$ as a function in one variable and optimizing with the method taught in Calc I. Petrus, when I return, I will check back in here first thing! :D

 

[Petrus]

No, you want to take the partial derivatives of $g(x,y)$ with respect to the two variables. What you should find is:

$$g_x(x,y)=g_y(x,y)=1$$

Do you see why?

Note: I must leave for about 4.5 hours now, so if anyone else wishes to jump in that is fine with me. Petrus has informed me that he has not actually studied Lagrange multipliers yet, but he looked it up and wants to use it here. I have also told him once this is completed, I would help him also finish the problem by using the constraint to express $f$ as a function in one variable and optimizing with the method taught in Calc I. Petrus, when I return, I will check back in here first thing! :D

Hello Mark,
I see why, I just confused myself exemple when you derivate respect to x I did treated y as a constant and forgot to treat 6 as a constant aswell, my bad.

 

[Jameson]

So now that you know $$g_x(x,y)=g_y(x,y)=1$$ you can set up the equations you need to find $\lambda$. MarkFL alright wrote out $f_x(x,y)$ and $f_y(x,y)$ so you don't have to do as much work to use this method.

$$x^4y^4e^{-3(x+y)}(5-3x)=1*\lambda$$

$$x^5y^3e^{-3(x+y)}(4-3y)=1*\lambda$$

$$x+y=6$$

Since the first two equations are equal to lambda, set them equal to each other and a bunch of stuff should cancel out. Once you simplify that you should have have an equation in terms of $x$ and $y$. Now use this and the third equation to find $x$ and $y$.

 

[MarkFL]

I was able to get in a few minutes here...(Whew)

Yes, what I did was what Jameson suggests...I equated the two first partials, simplified and solved for $y$, then plugged that into the constraint and obtained a quadratic in $x$. As I mentioned before, only one of the roots is $0<x<6$.

Can you find this valid root?

 

[Petrus]

I was able to get in a few minutes here...(Whew)

Yes, what I did was what Jameson suggests...I equated the two first partials, simplified and solved for $y$, then plugged that into the constraint and obtained a quadratic in $x$. As I mentioned before, only one of the roots is $0<x<6$.

Can you find this valid root?

Wat I get is $$(\frac{-10}{3},\frac{8}{3})$$ but that is not in the range... (have I done something wrong?)

 

[Jameson]

Wat I get is $$(\frac{-10}{3},\frac{8}{3}$$ but that is not in the range... (have I done something wrong?

Yes. It's going to be impossible to know what you did without seeing your work.

$$x^4y^4e^{-3(x+y)}(5-3x)=x^5y^3e^{-3(x+y)}(4-3y)$$

$$y(5-3x)=x(4-3y)$$

$$5y-3xy=4x-3xy \implies 5y-3xy+3xy=4x \implies 5y=4x$$

You can now solve for $y$ and then plug in that into $x+y=6$ to solve. Hopefully I didn't make any errors but check my work.

 

[Petrus]

Yes. It's going to be impossible to know what you did without seeing your work.

$$x^4y^4e^{-3(x+y)}(5-3x)=x^5y^3e^{-3(x+y)}(4-3y)$$

$$y(5-3x)=x(4-3y)$$

$$5y-3xy=4x-12xy \implies 5y-3xy+12xy=4x \implies y(5-3x+12x)=4x$$

You can now solve for $y$ and then plug in that into $x+y=6$ to solve. Hopefully I didn't make any errors but check my work.

Ok I am confused..? I get same as you but I get $$x(4-3y)=4x-3xy$$

 

[Jameson]

Ok I am confused..? I get same as you but I get $$x(4-3y)=4x-3xy$$

That is an error on my part. Thanks for pointing it out! :) Will edit now.

EDIT: Fixed. You should get $$\left( +\frac{10}{3}, \frac{8}{3} \right)$$. Somehow you ended up with a negative. This is definitely inside the triangle so that is a good sign. MarkFL said that he found two solutions though and we've only found one, so I hope he can point out if we took a wrong step.

 

[Petrus]

$$(\frac{6}{5},\frac{24}{5})$$
That means max:0.005 and min:0
I am correct?

 

[Jameson]

I'm not sure where you got those values. I posted the solution to the two equations in my previous post. Assuming that we correctly found the critical point, we need to plug that into $f(x,y)$ to find it's value so we can compare it to the other critical points. Like I said though, we should wait for MarkFL to comment on how we solved the Lagrange multiplier equations. Once we make sure that we've solved for the last critical point correctly then you are almost done!

 

[Petrus]

I'm not sure where you got those values. I posted the solution to the two equations in my previous post. Assuming that we correctly found the critical point, we need to plug that into $f(x,y)$ to find it's value so we can compare it to the other critical points. Like I said though, we should wait for MarkFL to comment on how we solved the Lagrange multiplier equations. Once we make sure that we've solved for the last critical point correctly then you are almost done!

Hello Jameson,
You got correct... I was just confusing with another problem I just finish ( It was a square a lot more easy:D), I was just rushing myself:P We wait for Mark:)

 

[MarkFL]

...MarkFL said that he found two solutions though and we've only found one, so I hope he can point out if we took a wrong step.

I made a silly blunder in my algebra. (Blush) Jameson is absolutely correct that we should in fact find:

$$y=\frac{5}{4}x$$

which means:

$$(x,y)=\left(\frac{10}{3},\frac{8}{3} \right)$$

Now, you want to determine which is greater, $$f\left(\frac{5}{3},\frac{4}{3} \right)$$ or $$f\left(\frac{10}{3},\frac{8}{3} \right)$$.

 

[Petrus]

I made a silly blunder in my algebra. (Blush) Jameson is absolutely correct that we should in fact find:

$$y=\frac{5}{4}x$$

which means:

$$(x,y)=\left(\frac{10}{3},\frac{8}{3} \right)$$

Now, you want to determine which is greater, $$f\left(\frac{5}{3},\frac{4}{3} \right)$$ or $$f\left(\frac{10}{3},\frac{8}{3} \right)$$.

Hi Mark
I now correctly find $$f\left(\frac{10}{3},\frac{8}{3} \right)$$ but how do you get $$f\left(\frac{5}{3},\frac{4}{3} \right)$$ from this equation? Or I am missunderstanding and there should be only one root?

 

[MarkFL]

The first point is inside the triangle (look at post #17, remember the critical point from equating the first partials to zero?) while the second point is on the boundary, specifically the hypotenuse. These are your two candidates for the absolute maximum, so you need to compare the function's value at those points and take the greater value as the absolute maximum.

 

[Petrus]

The first point is inside the triangle (look at post #17, remember the critical point from equating the first partials to zero?) while the second point is on the boundary, specifically the hypotenuse. These are your two candidates for the absolute maximum, so you need to compare the function's value at those points and take the greater value as the absolute maximum.

Yeah I know that but Then Jameson Said that you Said that you did find Two critical point from the last equation but $$f(\frac{10}{3},\frac{8}{3})<f(\frac{5}{3},\frac{4}{3})$$

 

[MarkFL]

I was mistaken, that was the silly blunder to which I referred. There is only 1 critical point on the hypotenuse as correctly found by Jameson.

Yes, you have correctly found that the function's value at the point in the interior is greater than the function's value at the point on the hypotenuse.

So, you are done! :D

 

[Petrus]

I was mistaken, that was the silly blunder to which I referred. There is only 1 critical point on the hypotenuse as correctly found by Jameson.

Yes, you have correctly found that the function's value at the point in the interior is greater than the function's value at the point on the hypotenuse.

So, you are done! :D

That means min value:$$f(0,0),f(x,0),f(0,y)$$ where x and y $$[0,6]$$
max value: $$f(\frac{5}{3},\frac{4}{3})$$
I am correct?