<s>[INST] If $a,b$ are positive integers and $x^2+y^2\leq 1$ then find the maximum of $ax+by$ without differentiation. If $x^2+y^2\leq 1$ then maximum of $ax+by$
Here what I have done so far.
Let $ax+by=k$ . Thus $by=k-ax$.
So we can have that $$b^2x^2+(k-ax)^2 \leq b^2$$
$$b^2x^2+k^2-2akx +a^2x^2-b^2\leq 0 $$
By re-writing as a quadratic of $x$ ,
$$(a^2+b^2)x^2-2akx +k^2-b^2\leq 0 $$
Since $a^2+b^2$ is positive , above quadratic has a minimum. Thus it to be negeative it must have roots.
So $$(-2ak)^2-4(a^2+b^2)(k^2-b^2) \geq 0$$
$$a^2k^2-(a^2+b^2)k^2+(a^2+b^2)b^2 \geq 0$$
$$(a^2+b^2) \geq k^2$$
So maximum of $k$ is $\sqrt{a^2+b^2}$
Is this correct ? 
If it is correct any shorter method ? Thanks in advance.
 [/INST] Let $x=r \cos \theta$ and $y=r \sin \theta$ with $r \leq 1$ so that $x^2+y^2 \leq 1$. Then,
$$ax+by=r(a\cos \theta+b\sin \theta)$$
But one may show that we may write,
$$a \cos \theta+b \sin \theta=\sqrt{a^2+b^2} \cos (\theta+\phi)$$
For some $\phi$.
Hence $ax+by \leq (1)(\sqrt{a^2+b^2})$.
</s>