Let $a$ and $b$ be positive real numbers. If $x^2 + y^2 \le1$ then the largest $ax + by$ is?

Question

The problem:

The solution:

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-Why is the equation $y=\frac{a}{b}x+\frac{c}{b}$ a line if $\frac{c}{b}$ is not constant? That is, $c$ varies as either $y$ or $x$ varies, but if this equation is a line, then the constant term $\frac{c}{b}$ can't vary, precisely because it's constant, which is not possible if $c$ is varying. Clearly, I'm missing a detail.

-A tangent line to a disc isn't necessary perpendicular to the line segment formed by the center of the disc and the point of interception between the line and the disc. The tangent line can be tilted up or down (so not perpendicular) and still be ''touching'' the disc, so why in this case this line is necessarily perpendicular?

Answer 1

"Why is the equation y=(a/b)x+c/b a line if c/b is not constant? That is, c varies as either y or x varies"

$c$ is a constant. The text kind of blew this. And then the abused this. Perhaps a clearer way of putting it would be:

Consider a value $c$ and all the $x,y$ such that $ax + by = c$. These points form a line. If this line passes through the interior of the disk then $c$ is a possible value for $ax + by$ where $x^2 + y^2 \le 1$. If this line misses the disk altogether then $c$ is not a possible value.

Different values of $c$ will create different lines, all parallel. larger values of $c$ will be to the "right" of lower values of $c$. The largest possible value of $c$ which will form the line that intersects the circle at just the furthest most right point that a parallel line with that slope can.

In other words the tangent line.

"A tangent line to a disc isn't necessary perpendicular to the line segment formed by the center of the disc and the point of interception between the line and the disc."

Actually, yes it is. That is a necessary condition.

Consider a line and a point not on the line. We can construct a line from the point to the line that is perpendicular to that line. Such a perpendicular is unique and length of this perpendicular is the shortest distance from the point to the line[$*$].

So if the tangent line is not perpendicular to the radius, then it is not the shortest distance from the center to line. So there is a point of the tangent line that is a shorter distance from the center than the radius. That would mean the tangent line passes through the interior of the circle and the tangent line is a chord. Presumably that is impossible as chords intersect a circle twice and tangent lines only once.

Furthermore, at a point of a circle we can create a line perpendicular to the radius. The radius will be the shortest distance from the center to the line so all other points of the line are further away from the center than the radius. Hence this line intersects the circle at only that on point.

So if tangent means "line that intersects the circle at exactly one point" then that line is precisely the perpendicular to the radius and no other such line exists.

If tangent means "line that intersects at a point and has the same slope of curve" then we have to prove that chords that intersect the circle at two points can not be tangent. That's intuitively obvious but I think we need calculus concepts to prove it.

[$*$] This is basically Euclid's fifth postulate consequences. A parallel line exists through the point, $p$, and a unique perpendicular exists at that point. The perpendicular crosses the original parallel line at a perpendicular as well at point $q$. If we take any other point, $s$ on the line we will have a right triangle $pqs$ and the hypotenuse $ps$ will be longer than the base $pq$. So $q$ is the unique point on the line that is closest to $p$.

Answer 2

The line $ax+by=c$ is the locus of points in the plane where the function $f(x,y)=ax+by$ attains the value $c$. Varying $c$, we have a family of parallel lines. We are looking for the largest $c$ such that the line $ax+by=c$ intersects the disc $x^2+y^2\leq 1$.

Answer 3

Alternatively, observe the following popular inequality: $(ax+by)^2 \le (a^2+b^2)(x^2+y^2)\le a^2+b^2 \implies |ax+by| \le \sqrt{a^2+b^2}\implies ax+by \le \sqrt{a^2+b^2}$, and this is the maximum that you sought.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\pars{a \quad b}{x \choose y} \leq \root{a^{2} + b^{2}}\root{x^{2} + y^{2}} \leq \root{a^{2} + b^{2}} \end{align}