Can someone show me the steps to solving these...
f(x,y) = x ln y + y ln x
df/dx =
df/dy =
df/dx^2 =
df/dy^2 =
df/dydx =
df/dxdy =
I believe df/dx = ln y + y/x
and df/dy = x/y + lnx
how would I find the critical points from that?
df/dx is the partial derivative of f(x, y) with respect to x; every other variable is considered a constant, so let's derive:
$\displaystyle \frac{df}{dx} = \ln{y}*x'+ y*(\ln{x})'$
Derivative w.r.t x of x is just 1. The derivative of ln(x) is 1/x
$\displaystyle \frac{df}{dx} = \ln{y} + \frac{y}{x}$
That simple? Yep. I believe you can solve the rest of the problem.
Finding the critical points is not easy.
I don't think it can be solved analytically(I could be wrong)
I drew a graph to see what it looked like and here it is.
So we want to solve the system of equations
$\displaystyle \ln(y)+\frac{y}{x}=0 \iff \frac{x}{y}=-\frac{1}{\ln(y)}$
$\displaystyle \frac{x}{y}+\ln(x)=0$ subbing the first into the 2nd we get
$\displaystyle -\frac{1}{\ln(y)}+\ln(x)=0 \iff -1 +\ln(x)\ln(y)=0 \iff ln(x)\ln(y)=1$
finally we get $\displaystyle \ln(y)=\frac{1}{\ln(x)} \iff y=e^{\frac{1}{\ln(x)}}$
If we sub this back into the first equation we get
$\displaystyle \frac{1}{\ln(x)}+\frac{e^{\frac{1}{\ln(x)}}}{x}=0$
Finally we can use newtons method on the last equation
$\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$
Based on the graph from above I used $\displaystyle x_0=0.5$
After applying newtons method three times I got the values(on a calculator)
$\displaystyle 0.5,0.3628837,0.36784455,0.36787943$
Oddly enough when I repeated the process but eliminted the x's and used newtons method again I got the same value for y.
so the critical numbers are $\displaystyle (\approx 0.36787943,\approx 0.36787943 $)
I hope this helps.
P.S if anyone knows another way I would love to see it
Thanks, The Empty Set.
I have been thinking about this and just noticed that
$\displaystyle \ln(y)+\frac{y}{x}=0 \iff x=-\frac{y}{\ln(y)}$ and
$\displaystyle \frac{x}{y}+\ln(x)=0 \iff y=-\frac{x}{\ln(x)}$
Are inverses of each other. So they are symmetric over the line y=x
This means that any of their intersections would have to be on the line y=x.
So if I sub this into the first equation I get
$\displaystyle x=-\frac{x}{\ln(x)} \iff \ln(x)=-1 \iff x=e^{-1}=\frac{1}{e} \approx 0.36787$
So the exact value is $\displaystyle x=y=\frac{1}{e}$