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Chandragupta Maurya

Let AP and BQ...

Question

Let $A P$ and $B Q$ be two vertical poles at points $A$ and $B,$ respectively. If $A P = 16 m, B Q = 22 m$ and $A B = 20 m,$ then find the distance of a point $R$ on $A B$ from the point $A$ such that $R P^{2} + R Q^{2}$ is minimum.

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Solution

Given data: $A P = 16 m, B Q = 22 m$ and $A B = 20 m .$
Drawing the figure, we get

Assume $A R = x, R B = 20 - x$
Using Pythagoras theorem, In $Δ A R P$
$R P^{2} = x^{2} + 16^{2} \dots (i)$
Again, Using Pythagoras theorem, In $Δ B R Q,$
$R Q^{2} = (20 - x)^{2} + 22^{2} \dots (i i)$
Now, let
$y = R P^{2} + R Q^{2}$
$= x^{2} + 16^{2} + (20 - x)^{2} + 22^{2}$
$= x^{2} + 256 + x^{2} - 40 x + 400 + 484$
$= 2 x^{2} - 40 x + 1140$
Differentiating w.r.t. $x$ ,
we get

$\frac{d y}{d x} = 4 x - 40$
$\Rightarrow \frac{d y}{d x} = 4 (x - 10)$
Putting, $\frac{d y}{d x} = 0$
$4 (x - 10) = 0$
$\Rightarrow x = 10$
Again differentiating, we get
$\frac{d^{2} y}{d x^{2}} = 4 > 0$

Using second derivative test $x = 10$ is point of minima.
Thus, the distance of $R$ from $A$ on $A B$ is $A R = x = 10 m .$

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Chandragupta Maurya

Standard IX History

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Let AP and BQ be two vertical poles at points A and B, respectively. If AP=16 m,BQ=22 m and AB=20 m, then find the distance of a point R on AB from the point A such that RP2+RQ2 is minimum.

Let $A P$ and $B Q$ be two vertical poles at points $A$ and $B,$ respectively. If $A P = 16 m, B Q = 22 m$ and $A B = 20 m,$ then find the distance of a point $R$ on $A B$ from the point $A$ such that $R P^{2} + R Q^{2}$ is minimum.