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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.

$${Let}\:{distance}\:{travelled}\:{be}\:{D}={d}_{\mathrm{1}} +{d}_{\mathrm{2}} . \\ $$$${Let}\:{x}\:{be}\:{the}\:{distance}\:{downstream}\: \\ $$$${where}\:{the}\:{man}\:{takes}\:{water}\:{from}\:{the} \\ $$$${river}\:{to}\:{the}\:{tent}.\:{Assuming}\:{the}\:{man}\:{runs}\:{along} \\ $$$${a}\:{straight}\:{line},\:{form}\:{a}\:{triangle}\:{ABC} \\ $$$${where}\:{AB}=\mathrm{200}\:{ft},{BC}=\left(\mathrm{450}−{x}\right){ft}, \\ $$$${AC}={d}_{\mathrm{1}} \:{and}\:{BC}\bot{AB}.\:{Thus},\:{by}\:{the}\:{Phythagorean}\:{theorem}, \\ $$$${d}_{\mathrm{1}} =\sqrt{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }. \\ $$$${In}\:{like}\:{manner},\:{d}_{\mathrm{2}} \:{is}\:{found}\:{to}\:{be}\: \\ $$$${d}_{\mathrm{2}} =\sqrt{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} }. \\ $$$$\therefore\:{D}=\sqrt{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }+\sqrt{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} } \\ $$$$\therefore\:\frac{{dD}}{{dx}}=\frac{−\mathrm{0}.\mathrm{5}\left(\mathrm{2}\left(\mathrm{450}−{x}\right)\right)}{\sqrt{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }}+\frac{{x}}{\sqrt{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} }} \\ $$$$\frac{{dD}}{{dx}}=\frac{\sqrt{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} }\left({x}−\mathrm{450}\right)+{x}\sqrt{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }}{\sqrt{\left(\mathrm{200}^{\mathrm{2}} +\left(\mathrm{450}−{x}\right)^{\mathrm{2}} \right)\left(\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)}} \\ $$$${At}\:{stationary}\:{points},\:\frac{{dD}}{{dx}}=\mathrm{0}. \\ $$$$\therefore\:{x}\sqrt{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }=\left(\mathrm{450}−{x}\right)\sqrt{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} } \\ $$$$\sqrt{\frac{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} }}=\frac{\mathrm{450}−{x}}{{x}} \\ $$$$\therefore\frac{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} }{\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\frac{\left(\mathrm{450}−{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} \left(\mathrm{450}−{x}\right)^{\mathrm{2}} +\mathrm{200}^{\mathrm{2}} {x}^{\mathrm{2}} =\left({x}−\mathrm{450}\right)^{\mathrm{2}} \left(\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} \right) \\ $$$$\left({x}−\mathrm{450}\right)^{\mathrm{2}} \left(\mathrm{100}^{\mathrm{2}} +{x}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)−\mathrm{200}^{\mathrm{2}} {x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left\{\mathrm{100}\left({x}−\mathrm{450}\right)\right\}^{\mathrm{2}} −\left(\mathrm{200}{x}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{100}{x}−\mathrm{45000}−\mathrm{200}{x}\right)\left(\mathrm{100}{x}−\mathrm{45000}+\mathrm{200}{x}\right)=\mathrm{0} \\ $$$$\left(\mathrm{100}{x}+\mathrm{45000}\right)\left(\mathrm{300}{x}−\mathrm{45000}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{450}\right)\left({x}−\mathrm{150}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{150}\:{ft}\:{or}\:{x}=−\mathrm{450}\:{ft}. \\ $$$${But}\:{x}>\mathrm{0}.\:\therefore\:{x}\neq−\mathrm{450}{ft}\:{and}\:{x}=\mathrm{150}{ft}. \\ $$$${Let}\:{f}\left({x}\right)=\frac{{dD}}{{dx}}. \\ $$$${Then}\:{f}\left(\mathrm{149}.\mathrm{9}\right)\approx−\mathrm{2}.\mathrm{56}×\mathrm{10}^{−\mathrm{4}} \\ $$$${and}\:{f}\left(\mathrm{150}.\mathrm{1}\right)\approx\mathrm{2}.\mathrm{56}×\mathrm{10}^{−\mathrm{4}} . \\ $$$${Hence}\:{f}\left({x}\right)\:{increases}\:{for}\:{increasing} \\ $$$${x}\:{sufficiently}\:{close}\:{to}\:{x}=\mathrm{150}.\:{Thus} \\ $$$${at}\:{x}=\mathrm{150}\:{ft},\:{D}\:{takes}\:{a}\:{minimum}\:{value}. \\ $$

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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