Let Ad and BC b e two vertical poles at A and B respectively on a horizontal ground. If AD = 8m, BC = 11 m and AB = 10 m, then the distance (in metres) of a point M on AB from the point A such that $MD^2+MC^2$ is minimum , is

5

The correct answer is option (2) : 5

Let the point M be at a distance x from the pole AD. Then,BM = $10 -x.$

$∴MD^2 =8^2 + x^2 $ and $MC^2 = (10-x)^2 + 11^2 $

Let $Z+ MD^2 +MC^2 .$ Then,

$Z= 8^2 + x^2 + (10 -x)^2 + 11^2 = 285 - 20 x + 2x^2 $

$⇒\frac{dZ}{dx}= -20 x + 4x $ and $ \frac{d^2Z}{dx^2 } = 4 $

For maximum and minimum values of Z, we must have

$\frac{dZ}{dx}= 0 ⇒-20 + 4x =0 ⇒ x= 5 $

Clearly, $\frac{d^2Z}{dx^2 } = 4 > 0 \, ∀ \, x $

Hence, Z is minimum when x = 5.