<s>[INST] Find maximum value of $\frac xy$ 
If $x^2-30x+y^2-40y+576=0$, find the maximum value of $\dfrac xy$.

First I completed the squares and got $(x-15)^2+(y-20)^2=7^2$, which is the equation of a circle.
I think I need to use some properties but I don't know what to do next.
 [/INST] Let $\frac{x}{y}=k$.
Thus, $x=ky$ and the equation
$$k^2y^2-30ky+y^2-40y+576=0$$ or
$$(k^2+1)y^2-2(15k+20)y+576=0$$ has real roots, which says that the discriminant must be non-negative.
Id est, $$(15k+20)^2-576(k^2+1)\geq0$$ or
$$351k^2-600k+176\leq0$$ or
$$\frac{44}{117}\leq k\leq\frac{4}{3}.$$
The value $\frac{4}{3}$ occurs for $$y=\frac{15k+20}{k^2+1}=14.4$$ and $$x=14.4\cdot\frac{4}{3}=19.2,$$ which says that $\frac{4}{3}$ is a maximal value.
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