Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 19472 by Tinkutara last updated on 11/Aug/17

A fishing boat is anchored 9 km away from the nearest point on the shore. A messanger must be sent from the fishing boat to a camp, 15 km from the point on shore closest to boat. If the messanger can walk at a speed of 5 km per hour and can row at 4 km/h, determine the distance of that point (in km) from the shore, where he must land so as to reach the shore in least possible time.

$$\mathrm{A}\:\mathrm{fishing}\:\mathrm{boat}\:\mathrm{is}\:\mathrm{anchored}\:\mathrm{9}\:\mathrm{km}\:\mathrm{away} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{shore}.\:\mathrm{A} \\ $$$$\mathrm{messanger}\:\mathrm{must}\:\mathrm{be}\:\mathrm{sent}\:\mathrm{from}\:\mathrm{the}\:\mathrm{fishing} \\ $$$$\mathrm{boat}\:\mathrm{to}\:\mathrm{a}\:\mathrm{camp},\:\mathrm{15}\:\mathrm{km}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point} \\ $$$$\mathrm{on}\:\mathrm{shore}\:\mathrm{closest}\:\mathrm{to}\:\mathrm{boat}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{messanger} \\ $$$$\mathrm{can}\:\mathrm{walk}\:\mathrm{at}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{5}\:\mathrm{km}\:\mathrm{per}\:\mathrm{hour} \\ $$$$\mathrm{and}\:\mathrm{can}\:\mathrm{row}\:\mathrm{at}\:\mathrm{4}\:\mathrm{km}/\mathrm{h},\:\mathrm{determine}\:\mathrm{the} \\ $$$$\mathrm{distance}\:\mathrm{of}\:\mathrm{that}\:\mathrm{point}\:\left(\mathrm{in}\:\mathrm{km}\right)\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{shore},\:\mathrm{where}\:\mathrm{he}\:\mathrm{must}\:\mathrm{land}\:\mathrm{so}\:\mathrm{as}\:\mathrm{to} \\ $$$$\mathrm{reach}\:\mathrm{the}\:\mathrm{shore}\:\mathrm{in}\:\mathrm{least}\:\mathrm{possible}\:\mathrm{time}. \\ $$

Commented by Tinkutara last updated on 11/Aug/17

Commented by mrW1 last updated on 11/Aug/17

t=((15−x)/5)+((√(9^2 +x^2 ))/4) (dt/dx)=−(1/5)+(x/(4(√(9^2 +x^2 ))))=0 (x/(√(9^2 +x^2 )))=(4/5) (x^2 /(81+x^2 ))=((16)/(25)) x^2 =16×9 x=12

$$\mathrm{t}=\frac{\mathrm{15}−\mathrm{x}}{\mathrm{5}}+\frac{\sqrt{\mathrm{9}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }}{\mathrm{4}} \\ $$$$\frac{\mathrm{dt}}{\mathrm{dx}}=−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{x}}{\mathrm{4}\sqrt{\mathrm{9}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }}=\mathrm{0} \\ $$$$\frac{\mathrm{x}}{\sqrt{\mathrm{9}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{81}+\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{16}}{\mathrm{25}} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{16}×\mathrm{9} \\ $$$$\mathrm{x}=\mathrm{12} \\ $$

Commented by Tinkutara last updated on 12/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com