Question

A man starts from the point P(3,-3) and reaches the point Q(0,2) after touching the line 2x+y=7 at R. The least value of PR+RQ is

Answer 1

Given: Coordinates of the initial point $P$ are $(3,\:-3)$ and coordinates of the terminal point $Q$ are $(0,\:2)$.

To find: The least value of $PR+RQ$.

Solution:

Given that, the man starts from the point $P$ and reaches the point $Q$ after touching the straight line $2x+y=7$ at $R$.

We consider the coordinates of the point $R$ be $(p,\:q)$.

Then, $2p+q=7\quad\quad .....(i)$

Now, $PR=\sqrt{(p-3)^{2}+(q+3)^{2}}$

$\quad=\sqrt{(p-3)^{2}+(7-2p+3)^{2}}$, by $(i)$

$\quad=\sqrt{p^{2}-6p+9+4p^{2}-40p+100}$

$\quad=\sqrt{5p^{2}-46p+109}$

and $RQ=\sqrt{(p-0)^{2}+(q-2)^{2}}$

$\quad=\sqrt{p^{2}+(7-2p-2)^{2}}$, by $(i)$

$\quad=\sqrt{p^{2}+4p^{2}-20p+25}$

$\quad=\sqrt{5p^{2}-20p+25}$

$\therefore PR+RQ$

$=\sqrt{5p^{2}-46p+109}+\sqrt{5p^{2}-20p+25}\quad\quad .....(ii)$

We have to find the least value of $PR+RQ$ at the point $p$; to find that point, we differentiate both sides of $(ii)$ by $p$ and equate with zero $(0)$.

$\quad\frac{d}{dp}(PQ+QR)=0$

$\Rightarrow \frac{d}{dp}(\sqrt{5p^{2}-46p+109}+\sqrt{5p^{2}-20p+25})=0$

$\Rightarrow \frac{10p-46}{2\sqrt{5p^{2}-46p+109}}+\frac{10p+20}{2\sqrt{5p^{2}-20p+25}}=0$

$\Rightarrow \frac{5p-23}{\sqrt{5p^{2}-46p+109}}=-\frac{5p-10}{\sqrt{5p^{2}-20p+25}}$

Squaring both sides, we get:

$\quad \frac{25p^{2}-230p+529}{5p^{2}-46p+109}=\frac{25p^{2}-100p+100}{5p^{2}-20p+25}$

Dividing both sides by 5, we get:

$\quad \frac{25p^{2}-230p+529}{25p^{2}-230p+545}=\frac{25p^{2}-100p+100}{25p^{2}-100p+125}$

Using dividendo, we get:

$\quad \frac{25p^{2}-230p+529}{25p^{2}-230p+545-25p^{2}+230p-529}=\frac{25p^{2}-100p+100}{25p^{2}-100p+125-25p^{2}+100p-100}$

$\Rightarrow \frac{25p^{2}-230p+529}{16}=\frac{25p^{2}-100p+100}{25}$

$\Rightarrow 625p^{2}-5750p+13225=400p^{2}-1600p+1600$

$\Rightarrow 225p^{2}-4150p+11625=0$

$\Rightarrow 9p^{2}-166p+465=0$

$\Rightarrow 9p^{2}-135p-31p+465=0$

$\Rightarrow 9p(p-15)-31(p-15)=0$

$\Rightarrow (p-15)(9p-31)=0$

Either $p-15=0$ or, $9p-31=0$

$\implies p=15,\:\frac{31}{9}$

• At $p=15,\:\frac{d^{2}(PQ+QR)}{dp^{2}}<0$
• At $p=\frac{31}{9},\:\frac{d^{2}(PQ+QR)}{dp^{2}}>0$

$\therefore$ at $p=\frac{31}{9}$, $(PQ+QR)$ attains the minimum value.

$\therefore Min(PQ+QR)=7.071$

Answer:

$\quad$The least value of PQ + QR is 7.071.