I need help w/ maxima points, 2 variables: f(x,y)=x*log(y)-y*log(x)

[MisterMaths]

So for the past hour i been trying to solve this to no avail:
f(x,y)=x*log(y)-y*log(x)
What needs to be done is to find stationary points and then determine maximum ones.
First of i went through derivation and then i tried to use the hessian method but i was stuck at (1-x+x^2y)/xy. Help a college student out!

 

[Deleted member 4993]

So for the past hour i been trying to solve this to no avail:
f(x,y)=x*log(y)-y*log(x)
What needs to be done is to find stationary points and then determine maximum ones.
First of i went through derivation and then i tried to use the hessian method but i was stuck at (1-x+x^2y)/xy. Help a college student out!

Your answer does not make sense!
f = x* log_e(y) - y * log_e(x)

df/dx = log_e(y) - y/x

df/dy = x/y - log_e(x)

Now what did you do? How did you get there?

 

[MisterMaths]

Your answer does not make sense!
f = x* log_e(y) - y * log_e(x)

df/dx = log_e(y) - y/x

df/dy = x/y - log_e(x)

Now what did you do? How did you get there?

Using the hessian method as a condition of second order (sufficient condition): H=f'xx*f'yy - (f'xy)^2= the answer i found myself in.
As you probably already know this is a function of 2 variables, derivatives alone aren't sufficient, so i need to use the method above to figure out the maximum points of the function, which i think is a saddle point but i can't prove it.

 

[ksdhart2]

Using the hessian method as a condition of second order (sufficient condition): H=f'xx*f'yy - (f'xy)^2= the answer i found myself in.
As you probably already know this is a function of 2 variables, derivatives alone aren't sufficient, so i need to use the method above to figure out the maximum points of the function, which i think is a saddle point but i can't prove it.

I'm not sure how helpful working with just the determinant of the Hessian will be though, as you have to evaluate it at a specific point to know if it's a maximum or minimum. But let's go with it and see what comes of it. The partial derivatives we need are:

\(\displaystyle \dfrac{\partial f}{\partial x\partial x}=\dfrac{y}{x^2}\)

\(\displaystyle \dfrac{\partial f}{\partial y\partial y}=-\dfrac{x}{y^2}\)

\(\displaystyle \dfrac{\partial f}{\partial x\partial y}=\dfrac{1}{y}-\dfrac{1}{x}\)

So then the determinant of the Hessian is:

\(\displaystyle D=\dfrac{y}{x^2} \cdot -\dfrac{x}{y^2} - \dfrac{(x - y)^2}{x^2 \cdot y^2}=-\dfrac{xy}{x^2 y^2} - \dfrac{(x - y)^2}{x^2 y^2}=\dfrac{-x^2+xy-y^2}{x^2 y^2}\)

This really tells us nothing without any points to check it against. So now we need to go back to the first derivatives and set them to zero. What happens when you do that?

 

[MisterMaths]

Setting the first derivates to equal zero as in when the rate of change is null, i get the following:
f'(x)=log(y)-y/x=0
f'(y)=x/y-log(x)=0
Log(x)=y/x and Log(y)=x/y

And from here it's unclear how to determine extrema points.

 

[Deleted member 4993]

Setting the first derivates to equal zero as in when the rate of change is null, i get the following:
f'(x)=log(y)-y/x=0
f'(y)=x/y-log(x)=0
Log(x)=y/x and Log(y)=x/y

And from here it's unclear how to determine extrema points.

You can observe that the above equations are satisfied at x = y = e

 

[ksdhart2]

Setting the first derivates to equal zero as in when the rate of change is null, i get the following:
f'(x)=log(y)-y/x=0
f'(y)=x/y-log(x)=0
Log(x)=y/x and Log(y)=x/y

And from here it's unclear how to determine extrema points.

Let's rearrange the terms a bit:

\(\displaystyle ln(x)=\dfrac{y}{x} \implies y=xln(x)\) and \(\displaystyle ln(y)=\dfrac{x}{y} \implies x=yln(y)\).

Then this would be a good time to back substitute. Let's plug in y=xln(x) into f'(x):

\(\displaystyle ln(y)-\dfrac{y}{x}=0 \implies ln(xln(x))-\dfrac{xln(x)}{x}=0 \implies ln(xln(x)) - ln(x)=0\)

Applying a logarithm rule, we can then see that:

\(\displaystyle ln(xln(x)) - ln(x)=0 \implies ln \left(\dfrac{xln(x)}{x} \right) = 0 \implies ln(ln(x)) = 0 \implies ln(x)=e^0=1 \implies x=e^1=e\)

Similar logic can be applied to arrive at a value for y. What value is that? Returning to the Hessian, what can you conclude about any maxima or minima?

 

[MisterMaths]

Let's rearrange the terms a bit:

\(\displaystyle ln(x)=\dfrac{y}{x} \implies y=xln(x)\) and \(\displaystyle ln(y)=\dfrac{x}{y} \implies x=yln(y)\).

Then this would be a good time to back substitute. Let's plug in y=xln(x) into f'(x):

\(\displaystyle ln(y)-\dfrac{y}{x}=0 \implies ln(xln(x))-\dfrac{xln(x)}{x}=0 \implies ln(xln(x)) - ln(x)=0\)

Applying a logarithm rule, we can then see that:

\(\displaystyle ln(xln(x)) - ln(x)=0 \implies ln \left(\dfrac{xln(x)}{x} \right) = 0 \implies ln(ln(x)) = 0 \implies ln(x)=e^0=1 \implies x=e^1=e\)

Similar logic can be applied to arrive at a value for y. What value is that? Returning to the Hessian, what can you conclude about any maxima or minima?

I followed up on your way and rechecked and it seems there's a mistake in the initial form, The correct form is ln(x)=x/y not ln(x)=y/x, it's apparently my fault. In my last reply i mistakenly confused the values of ln(x) and ln(y).

 

[Deleted member 4993]

I followed up on your way and rechecked and it seems there's a mistake in the initial form, The correct form is ln(x)=x/y not ln(x)=y/x, it's apparently my fault. In my last reply i mistakenly confused the values of ln(x) and ln(y).

But you have followed his method (shown) and got the correct answer/s - right?

 

[MisterMaths]

Yes of course, thanks for the assistance!