Let $f(x, y)=ax+by+c$. Find the largest value of $r$ such that $f(x, y)>0$ for all pairs $(x, y)$ satisfying $x^2 + y^2 < r^2$?

Question

Let $f(x, y) = ax + by + c$, where $\{a, b, c\} \subset\mathbb R$ and $c > 0$. Find the largest value of $r$ such that $f(x, y) > 0$ for all pairs $(x, y)$ satisfying $x^2 + y^2 < r^2$.

I was trying this question many times, but I could not find it. I don't know where i have to start.

I was taking taking the radius $r= 1$, $x= \frac{1}{2}$ and $y =\frac{1}{2}$, possisbly it is meaningless to take this value because it never less than $1$ as $\frac{1}{2} + \frac{1}{2}= 1$ so it is contradict.

If anybody help me I would be very thankful to him.

Answer 1

By C-S $$r^2>x^2+y^2=\frac{(a^2+b^2)(x^2+y^2)}{a^2+b^2}\geq\frac{(ax+by)^2}{a^2+b^2}.$$ Thus, $$-r\sqrt{a^2+b^2}<ax+by<r\sqrt{a^2+b^2}$$ or $$c-r\sqrt{a^2+b^2}<ax+by+c<c+r\sqrt{a^2+b^2}.$$ Thus, we need $$c-r\sqrt{a^2+b^2}\geq0$$ or $$r\leq\frac{c}{\sqrt{a^2+b^2}}.$$

Answer 2

Note that the graph of $z=ax+by+c$ is a plane that has $z$-intercept at $(0,0,c)$. Depending on the signs of the coefficients of $a,b$, there are four cases of $x,y$-intercepts at $(\frac{c}{a},0,0), (0,\frac{c}{b},0)$.

The constraint is a circle with the radius $r$ and center at the origin, which ($r$) must be less than the distance to the line passing through the $x,y$-intercepts, so that the plane is above the $XY$ plane (i.e. $f(x,y)>0$). Thus: $$r<\frac{\bigg{|} \frac{c}{a} \bigg{|} \cdot \bigg{|} \frac{c}{b} \bigg{|} }{\sqrt{\left(\frac{c}{a}\right)^2+\left(\frac{c}{b}\right)^2}}=\frac{c}{\sqrt{a^2+b^2}}.$$