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Question Number 5630 by Rasheed Soomro last updated on 23/May/16

Commented by Rasheed Soomro last updated on 23/May/16

Commented by Yozzii last updated on 23/May/16
$Let distance travelled be D=d_1 +d_2 . Let x be the distance downstream where the man takes water from the river to the tent. Assuming the man runs along a straight line, form a triangle ABC where AB=200 ft,BC=(450−x)ft, AC=d_1 and BC⊥AB. Thus, by the Phythagorean theorem, d_1 =(√((450−x)^2 +200^2 )). In like manner, d_2 is found to be d_2 =(√(100^2 +x^2 )). ∴ D=(√((450−x)^2 +200^2 ))+(√(100^2 +x^2 )) ∴ (dD/dx)=((−0.5(2(450−x)))/(√((450−x)^2 +200^2 )))+(x/(√(100^2 +x^2 ))) (dD/dx)=(((√(100^2 +x^2 ))(x−450)+x(√((450−x)^2 +200^2 )))/(√((200^2 +(450−x)^2 )(100^2 +x^2 )))) At stationary points, (dD/dx)=0. ∴ x(√((450−x)^2 +200^2 ))=(450−x)(√(100^2 +x^2 )) (√(((450−x)^2 +200^2 )/(100^2 +x^2 )))=((450−x)/x) ∴(((450−x)^2 +200^2 )/(100^2 +x^2 ))=(((450−x)^2 )/x^2 ) x^2 (450−x)^2 +200^2 x^2 =(x−450)^2 (100^2 +x^2 ) (x−450)^2 (100^2 +x^2 −x^2 )−200^2 x^2 =0 {100(x−450)}^2 −(200x)^2 =0 (100x−45000−200x)(100x−45000+200x)=0 (100x+45000)(300x−45000)=0 ⇒(x+450)(x−150)=0 ⇒x=150 ft or x=−450 ft. But x>0. ∴ x≠−450ft and x=150ft. Let f(x)=(dD/dx). Then f(149.9)≈−2.56×10^(−4) and f(150.1)≈2.56×10^(−4) . Hence f(x) increases for increasing x sufficiently close to x=150. Thus at x=150 ft, D takes a minimum value.$
$L e t d i s t a n c e t r a v e l l e d b e D = d_{1} + d_{2} .$ $L e t x b e t h e d i s t a n c e d o w n s t r e a m$ $w h e r e t h e m a n t a k e s w a t e r f r o m t h e$ $r i v e r t o t h e t e n t . A s s u m i n g t h e m a n r u n s a l o n g$ $a s t r a i g h t l i n e, f o r m a t r i a n g l e A B C$ $w h e r e A B = 200 f t, B C = (450 - x) f t,$ $A C = d_{1} a n d B C ⊥ A B . T h u s, b y t h e P h y t h a g o r e a n t h e o r e m,$ $d_{1} = \sqrt{{(450 - x)}^{2} + 200^{2}} .$ $I n l i k e m a n n e r, d_{2} i s f o u n d t o b e$ $d_{2} = \sqrt{100^{2} + x^{2}} .$ $∴ D = \sqrt{{(450 - x)}^{2} + 200^{2}} + \sqrt{100^{2} + x^{2}}$ $∴ \frac{d D}{d x} = \frac{- 0.5 (2 (450 - x))}{\sqrt{{(450 - x)}^{2} + 200^{2}}} + \frac{x}{\sqrt{100^{2} + x^{2}}}$ $\frac{d D}{d x} = \frac{\sqrt{100^{2} + x^{2}} (x - 450) + x \sqrt{{(450 - x)}^{2} + 200^{2}}}{\sqrt{(200^{2} + {(450 - x)}^{2}) (100^{2} + x^{2})}}$ $A t s t a t i o n a r y p o i n t s, \frac{d D}{d x} = 0 .$ $∴ x \sqrt{{(450 - x)}^{2} + 200^{2}} = (450 - x) \sqrt{100^{2} + x^{2}}$ $\sqrt{\frac{{(450 - x)}^{2} + 200^{2}}{100^{2} + x^{2}}} = \frac{450 - x}{x}$ $∴ \frac{{(450 - x)}^{2} + 200^{2}}{100^{2} + x^{2}} = \frac{{(450 - x)}^{2}}{x^{2}}$ $x^{2} {(450 - x)}^{2} + 200^{2} x^{2} = {(x - 450)}^{2} (100^{2} + x^{2})$ ${(x - 450)}^{2} (100^{2} + x^{2} - x^{2}) - 200^{2} x^{2} = 0$ ${100 (x - 450)}^{2} - {(200 x)}^{2} = 0$ $(100 x - 45000 - 200 x) (100 x - 45000 + 200 x) = 0$ $(100 x + 45000) (300 x - 45000) = 0$ $\Rightarrow (x + 450) (x - 150) = 0$ $\Rightarrow x = 150 f t o r x = - 450 f t .$ $B u t x > 0 . ∴ x \neq - 450 f t a n d x = 150 f t .$ $L e t f (x) = \frac{d D}{d x} .$ $T h e n f (149.9) \approx - 2.56 \times 10^{- 4}$ $a n d f (150.1) \approx 2.56 \times 10^{- 4} .$ $H e n c e f (x) i n c r e a s e s f o r i n c r e a s i n g$ $x s u f f i c i e n t l y c l o s e t o x = 150 . T h u s$ $a t x = 150 f t, D t a k e s a m i n i m u m v a l u e .$
Commented by Rasheed Soomro last updated on 23/May/16

$The value of x you calculated is correct,$ $but the answer is 450 - x = 300 ft upstream$ $from the location of hiker .$ $Your aproach is different from that of the$ $book .$ $According to the book, the point of the river$ $to which hiker should run in order to minimize$ $the distance is such that two triangles made are$ $similar .$ $Anyway I Like your aproach!$
Commented by Yozzii last updated on 23/May/16

$D o y o u h a v e a p r o o f o f t h e a s s e r t i o n$ $t h a t t h e s u m d_{1} + d_{2} i s m i n i m i s e d$ $i f t h e t w o r i g h t - a n g l e d t r i a n g l e s a r e s i m i l a r ?$
Commented by Rasheed Soomro last updated on 23/May/16

$I am forwarding the explanation of$ $the book regarding your request .$
Commented by Rasheed Soomro last updated on 23/May/16

Commented by Rasheed Soomro last updated on 23/May/16

Commented by Yozzii last updated on 23/May/16

$I s e e . T h a n k s!$
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