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Question-5630




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.
$${Let}\:{distance}\:{travelled}\:{be}\:{D}={d}_{\mathrm{1}} +{d}_{\mathrm{2}} . \\ $$$${Let}\:{x}\:{be}\:{the}\:{distance}\:{downstream}\: \\ $$$${where}\:{the}\:{man}\:{takes}\:{water}\:{from}\:{the} \\ $$$${river}\:{to}\:{the}\:{tent}.\:{Assuming}\:{the}\:{man}\:{runs}\:{along} \\ $$$${a}\:{straight}\:{line},\:{form}\:{a}\:{triangle}\:{ABC} \\ $$$${where}\:{AB}=\mathrm{200}\:{ft},{BC}=\left(\mathrm{450}−{x}\right){ft}, \\ $$$${AC}={d}_{\mathrm{1}} \:{and}\:{BC}\bot{AB}.\:{Thus},\:{by}\:{the}\:{Phythagorean}\:{theorem}, \\ $$$${d}_{\mathrm{1}} =\sqrt{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }. \\ $$$${In}\:{like}\:{manner},\:{d}_{\mathrm{2}} \:{is}\:{found}\:{to}\:{be}\: \\ $$$${d}_{\mathrm{2}} =\sqrt{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} }. \\ $$$$\therefore\:{D}=\sqrt{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }+\sqrt{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} } \\ $$$$\therefore\:\frac{{dD}}{{dx}}=\frac{−\mathrm{0}.\mathrm{5}\left(\mathrm{2}\left(\mathrm{450}−{x}\right)\right)}{\:\sqrt{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }}+\frac{{x}}{\:\sqrt{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} }} \\ $$$$\frac{{dD}}{{dx}}=\frac{\sqrt{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} }\left({x}−\mathrm{450}\right)+{x}\sqrt{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }}{\:\sqrt{\left(\mathrm{200}^{\mathrm{2}} +\left(\mathrm{450}−{x}\right)^{\mathrm{2}} \right)\left(\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)}} \\ $$$${At}\:{stationary}\:{points},\:\frac{{dD}}{{dx}}=\mathrm{0}. \\ $$$$\therefore\:{x}\sqrt{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }=\left(\mathrm{450}−{x}\right)\sqrt{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} } \\ $$$$\sqrt{\frac{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} }}=\frac{\mathrm{450}−{x}}{{x}} \\ $$$$\therefore\frac{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\frac{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} \left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} {x}^{\mathrm{2}} =\left({x}−\mathrm{450}\right)^{\mathrm{2}} \left(\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} \right) \\ $$$$\left({x}−\mathrm{450}\right)^{\mathrm{2}} \left(\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)−\mathrm{200}^{\mathrm{2}} {x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left\{\mathrm{100}\left({x}−\mathrm{450}\right)\right\}^{\mathrm{2}} −\left(\mathrm{200}{x}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{100}{x}−\mathrm{45000}−\mathrm{200}{x}\right)\left(\mathrm{100}{x}−\mathrm{45000}+\mathrm{200}{x}\right)=\mathrm{0} \\ $$$$\left(\mathrm{100}{x}+\mathrm{45000}\right)\left(\mathrm{300}{x}−\mathrm{45000}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{450}\right)\left({x}−\mathrm{150}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{150}\:{ft}\:{or}\:{x}=−\mathrm{450}\:{ft}. \\ $$$${But}\:{x}>\mathrm{0}.\:\therefore\:{x}\neq−\mathrm{450}{ft}\:{and}\:{x}=\mathrm{150}{ft}. \\ $$$${Let}\:{f}\left({x}\right)=\frac{{dD}}{{dx}}. \\ $$$${Then}\:{f}\left(\mathrm{149}.\mathrm{9}\right)\approx−\mathrm{2}.\mathrm{56}×\mathrm{10}^{−\mathrm{4}} \\ $$$${and}\:{f}\left(\mathrm{150}.\mathrm{1}\right)\approx\mathrm{2}.\mathrm{56}×\mathrm{10}^{−\mathrm{4}} . \\ $$$${Hence}\:{f}\left({x}\right)\:{increases}\:{for}\:{increasing} \\ $$$${x}\:{sufficiently}\:{close}\:{to}\:{x}=\mathrm{150}.\:{Thus} \\ $$$${at}\:{x}=\mathrm{150}\:{ft},\:{D}\:{takes}\:{a}\:{minimum}\:{value}. \\ $$
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!
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{x}\:\mathrm{you}\:\mathrm{calculated}\:\mathrm{is}\:\mathrm{correct}, \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{450}−\mathrm{x}=\mathrm{300}\:\mathrm{ft}\:\mathrm{upstream} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{location}\:\mathrm{of}\:\mathrm{hiker}. \\ $$$$\mathrm{Your}\:\mathrm{aproach}\:\mathrm{is}\:\mathrm{different}\:\mathrm{from}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{book}. \\ $$$$\mathrm{According}\:\mathrm{to}\:\mathrm{the}\:\mathrm{book},\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{river} \\ $$$$\mathrm{to}\:\mathrm{which}\:\mathrm{hiker}\:\:\mathrm{should}\:\mathrm{run}\:\mathrm{in}\:\mathrm{order}\:\mathrm{to}\:\mathrm{minimize} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{is}\:\mathrm{such}\:\mathrm{that}\:\mathrm{two}\:\mathrm{triangles}\:\mathrm{made}\:\mathrm{are} \\ $$$$\mathrm{similar}. \\ $$$$\mathrm{Anyway}\:\mathrm{I}\:\mathrm{Like}\:\mathrm{your}\:\mathrm{aproach}! \\ $$
Commented by Yozzii last updated on 23/May/16
Do you have a proof of the assertion  that the sum d_1 +d_2  is minimised  if the two right−angled triangles are similar?
$${Do}\:{you}\:{have}\:{a}\:{proof}\:{of}\:{the}\:{assertion} \\ $$$${that}\:{the}\:{sum}\:{d}_{\mathrm{1}} +{d}_{\mathrm{2}} \:{is}\:{minimised} \\ $$$${if}\:{the}\:{two}\:{right}−{angled}\:{triangles}\:{are}\:{similar}? \\ $$
Commented by Rasheed Soomro last updated on 23/May/16
I am forwarding the explanation of  the book regarding your request.
$$\mathrm{I}\:\mathrm{am}\:\mathrm{forwarding}\:\mathrm{the}\:\mathrm{explanation}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{book}\:\mathrm{regarding}\:\mathrm{your}\:\mathrm{request}. \\ $$
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 see. Thanks!
$${I}\:{see}.\:{Thanks}! \\ $$

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