If $a,b$ are positive integers and $x^2+y^2\leq 1$ then find the maximum of $ax+by$ without differentiation.

Use Cauchy Буняко́вський Schwarz, $|\langle (a,b), (x,y) \rangle| \le \|(a,b)\| \|x,y\|$. Choosing $(x,y)= {1 \over \|(a,b)\|} (a,b)$ shows equality.

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 $ax+by \leq (1)(\sqrt{a^2+b^2})$.

We can solve this by geometry,

$$x^2+y^2\le 1$$ is area bounded by circle with centre at origin and radius 1

&

$$ax+by=c$$ Is a line

We want to find max value of c,

Note: a and b are positive so slope is negative.

We want to find (x,y) such that it lies in circle and on line but also that line gives us maximum intersect at y-axis (which is $=\frac{c}{b}$)

If i draw the circle, and couple of lines

In above picture we see that y-intersect is max when line is tangent to circle

Using basic coordinate knowledge, we can deduce the web of line such that it is tangent to circle

and then get c.