A factory D is to be connected by a road with a straight railway line on which a town A is situated. The distance DB of the factory to the railway line is 53 km. Length AB of the railway line is 20km. Freight charges on the road are twice the charges on the railway. The point P(AP

Let then charges for railways line be k Rs /km Now total freight charges, T = k x + 2 k ( 20 − x ) 2 + 75 L e t d T d x = 0 O r k + 2 k 2 ( x − 20 ) 2 ( x − 20 ) 2 + 75 = 0 O r 4 ( x − 20 ) 2 = 75 + ( x − 20 ) 2 O r ( x − 20 ) 2 = 25 O r x = 25 , 15 , O r x = 15 ( a s A P < A B ) O r P B = A B − A P = 20 − 15 = 5 k m

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A factory D is to be connected by a road with a straight railway line on which a town A is situated. The distance DB of the factory to the railway line is $5 \sqrt{3}$ km. Length AB of the railway line is 20km. Freight charges on the road are twice the charges on the railway. The point $P (A P < A B)$ on the railway line should the road DP be connected so as to ensure minimum freight charges from the factory to the town is

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$B P = 7.5 k m$

$A P = 15$

$B P = 5 k m$

$A P = 5 k m$

answer is A, D.

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Detailed Solution

Let then charges for railways line be k Rs /km
Now total freight charges, $T = k x + 2 k \sqrt{{(20 - x)}^{2} + 75}$
$L e t \frac{d T}{d x} = 0$

$O r k + 2 k \frac{2 (x - 20)}{2 \sqrt{{(x - 20)}^{2} + 75}} = 0 O r 4 {(x - 20)}^{2} = 75 + {(x - 20)}^{2} O r {(x - 20)}^{2} = 25 O r x = 25, 15, O r x = 15 (a s A P < A B) O r P B = A B - A P = 20 - 15 = 5 k m$