finding the extreme values of $f$.

Question

If $f(x,y) = x ^2 + xy + y^2 - 4 \ln x - 10 \ln y,$ and I found $f_{x}^{'} = 2x + y - 4/x$ which means, $$\tag{1}2x^2 + xy = 4$$ and also I found $f_{y}^{'} = 2y + x - 10/y$ which means, $$\tag{2}2y^2 + xy = 10$$

then I subtracted (1) from (2) and I got $$y^2 - x^2 = 3,$$

But then I do not know how to complete, can anyone help me please?

Answer 1

You have:

• $2x^2+xy=4$
• $2y^2+xy=10$

Now solve for one variable (1st equation) to get $$y=\frac{4}{x}-2x$$ Plugging this into the 2nd equation yields: $$2\left(\frac{4}{x}-2x\right)^2+x\left(\frac{4}{x}-2x\right)=10$$ This further simpliefies to: $$3x^4-19x^2+16=0$$ From here one you can use the quadratic formula to get $x\in\{-\frac{4}{\sqrt 3},-1,1,\frac{4}{\sqrt 3}\}$. Only the positive numbers are valid solutions though.

Answer 2

Your problem is that you are missing one equation because you combined them. What you want to do is substitute the variables:

$$2x+y-\dfrac{4}{x} = 0 \Rightarrow y = \dfrac{4}{x} -2x$$

Now plugging this into the other equation:

$$2\left(\dfrac{4}{x}-2x\right)^2 + x \cdot \left( \dfrac{4}{x} -2x \right) = 10$$

Which can be solved for x.