Review question

# Given the line $PRQ$, when is $OP + OQ$ a minimum? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7586

## Solution

The line $L$ has equation $y=c-mx$, with $m>0$ and $c>0$. It passes through the point $R(a,b)$ and cuts the axes at the points $P(p,0)$ and $Q(0,q)$, where $a$, $b$, $p$ and $q$ are all positive. Find $p$ and $q$ in terms of $a$, $b$ and $m$.

The equation of a line with gradient $-m$, passing through the point $R(a,b)$ is $y-b=-m(x-a).$ We can rearrange this and compare it with the given equation for $L$ to find that $c= ma+b.$ Therefore we find that $q=ma+b$ and $p=\dfrac{ma+b}{m}$.

As $L$ varies with $R$ remaining fixed, show that the minimum value of the sum of the distances of $P$ and $Q$ from the origin is $(a^{\frac{1}{2}}+b^{\frac{1}{2}})^2$

Use the applet to explore the behaviour as $m$ varies. Drag $R$ to different locations and vary $m$ to see how the values of $OP+OQ$ and $PQ$ change.

Let the sum of the distances of $P$ and $Q$ from the origin be $D$. Then $D = (ma+b) + \left(\frac{ma+b}{m}\right) = ma+\frac{b}{m}+a+b.$ Since $a$ and $b$ are fixed, by varying $L$ we are effectively just varying $m$. Therefore, we want to minimise $D$ with respect to $m$.

Now differentiating, we have $\frac{dD}{d m}=a-\frac{b}{m^2}.$ We get a stationary point when $\dfrac{dD}{dm}=0$. That gives $a=\frac{b}{m^2} \Longleftrightarrow m=\pm \sqrt{\frac{b}{a}}.$ We are told that $m>0$, so we choose the positive root $m=\sqrt{\frac{b}{a}}$.

This value must be a minimum value as there is only one stationary value, and by considering our expression for $D$, we can see that as $m\rightarrow 0, \, D$ will become very large, and the same will happen as $m\rightarrow\infty$.

To find the minimum value, we substitute back into the expression for $D$: \begin{align*} D &= a+b+a\sqrt{\frac{b}{a}}+b\sqrt{\frac{a}{b}}\\ &= a+b+2a^{\frac{1}{2}}b^{\frac{1}{2}}\\ &= (a^{\frac{1}{2}}+b^{\frac{1}{2}})^2 \end{align*}

as required.

… and find in a similar form the minimum distance between $P$ and $Q$.

The distance $PQ = \sqrt{p^2+q^2}$.

It’s helpful for us to note that the minimum value of $\sqrt{p^2+q^2}$ coincides with the minimum value of $p^2+q^2$, so we can make our lives a little easier by concentrating on minimising $PQ^2 = p^2 + q^2 = \left(\frac{ma+b}{m}\right)^2 + (ma+b)^2$ rather than $\sqrt{p^2+q^2}$.

\begin{align*} PQ^2 &=\left(\frac{ma+b}{m}\right)^2 + (ma+b)^2 \\ &= \left(a+\frac{b}{m}\right)^2 + (ma + b)^2 \\ &=a^2 + 2\frac{ab}{m} + \frac{b^2}{m^2} + a^2m^2 + 2abm+ b^2\\ \end{align*}

Differentiating $PQ^2$ with respect to $m$ gives

$\frac{d}{dm}(PQ^2) =-\frac{2ab}{m^2}-\frac{2b^2}{m^3} + 2a^2m + 2ab$

The stationary points occur when

\begin{align*} -\frac{2ab}{m^2}-\frac{2b^2}{m^3} + 2a^2m + 2ab &= 0 \\ \iff \quad -\frac{b}{m^3}(am+b)+a(am+b)&=0 \\ \iff \quad \qquad (am + b)\left(a - \frac{b}{m^3}\right)&=0. \end{align*} The two (real) roots of this equation are \begin{align*} m&=-\frac{b}{a}\\ m^3&=\frac{b}{a} \quad \iff m=\sqrt[3]{\frac{b}{a}}. \end{align*} The first of these is not a valid root, as $m>0$, so substituting the second back into the expression for $PQ^2$ we find \begin{align*} p^2+q^2&=\left(1+\frac{1}{m^2}\right)(ma+b)^2\\ &=(1+a^{\frac{2}{3}}b^{-\frac{2}{3}})(a^{\frac{2}{3}}b^{\frac{1}{3}}+b)^2\\ &=b^{-\frac{2}{3}}(b^{\frac{2}{3}}+a^{\frac{2}{3}})(b^{\frac{1}{3}})^2(a^{\frac{2}{3}}+b^{\frac{2}{3}})^2\\ &=(a^{\frac{2}{3}}+b^{\frac{2}{3}})^3. \end{align*}

Thus the minimum value of $PQ= \sqrt{p^2+q^2}$ as $L$ varies is $(a^{\frac{2}{3}}+b^{\frac{2}{3}})^{\frac{3}{2}}.$

As before, this value must be a minimum value as there is only one legitimate stationary value, and we can create arbitrarily large $PQ$ by taking $m\rightarrow 0$ or $m\rightarrow\infty$.