How to solve a simple non-linear optimisation problem in order to find the minimum point in a hyperbola?

Question

Consider the equation

$$0.26639x-0.043941y+(5.9313\times10^{-5})xy-(3.9303\times{10^{-6}}) y^2-7242.0404=0$$

with $x,y>0$. If you plot it, it'll look like below:

Now, I want to find a minimum point on this hyperbola, such that $x+y$ is a minimum. In other words: $$\min(x+y)$$ $$Constraints: $$ $$0.26639x-0.043941y+(5.9313\times10^{-5})xy-(3.9303\times{10^{-6}}) y^2-7242.0404=0$$

Any help on how to mathematically find this point would be really helpful.

I've asked a similar question here, but in this one, I wanted to find a corner point such that the hyperbola has the maximum curvature. But this is not the case in this question.

Answer 1

Hint.

Assuming the plot gives the restriction shape, the minimum is located in the first quadrant, at the tangency point between the restriction and the line $x+y=\lambda$. Now calling the restriction

$$ g(x,y) = a x + b y + c x y + d y^2 + e = 0 $$

making the substitution $y = \lambda-x$ we get

$$ a x+b (\lambda -x)+c x (\lambda -x)+d (\lambda -x)^2+e = 0\ \ \ (1) $$

and after solving for $x$ we get

$$ x = \frac{2 d \lambda\pm \sqrt{(a-b+\lambda (c-2 d))^2+4 (c-d) (\lambda (b+d \lambda )+e)}-a+b-c \lambda }{2 (d-c)} $$

but at tangency we have only one solution for $x$ so

$$ (a-b+\lambda (c-2 d))^2+4 (c-d) (\lambda (b+d \lambda )+e)=0 $$

for $\lambda = 18287.7$ and after substituting into $(1)$

$$ x = 12165.6 $$

Answer 2

Our goal, as I mentioned in my first comment, is to find the line of the form $y=-x+c$ that minimizes $c,$ and which intersects your hyperbola in the first quadrant.

1. Find $y'(x).$ I'm actually going to view $x=x(y)$ and find $dx/dy.$ Note that if $dy/dx=-1,$ then $dx/dy=-1$ as well. Moreover, in the other post, I found $x(y)$ explicitly. We have: \begin{align*} x(y)&=\frac{7242.0404+\left(3.9303\times{10^{-6}}\right) y^2+0.043941y}{0.26639+\left(5.9313\times10^{-5}\right)\!y} \\ x'(y)&=\frac{0.0662637 y^2+595.215 y-1.18771\times 10^8}{(1. y+4491.26)^2} \end{align*}
2. Set $x'(y)=-1.$ We have $x'(y)=-1 \implies y=6122.12, \; x=12165.6.$ This solution is in the first quadrant, so we see that there is a point satisfying our needs. Note: this was also my first approach in the previous problem, but it didn't give quite as good a result for that problem.
3. Now, we only need find the $c$ that serves as the $y$ intersept, and that will be the minimum. We have \begin{align*} y&=-x+c \\ 6122.12&=-12165.6+c \\ 18287.7&=c. \end{align*} So that's the minimum, and it occurs at $(12165.6, 6122.12).$