The distance between cities X and Y is equal to 21 kilometers. A pedestrian leaves X and goes to Y at a constant speed of 5 kilometers per hour. At the same moment, a cyclist leaves Y and goes towards X, the speed of the latter can vary between 10 and 13 km/h throughout the journey. After meeting each other, the motorcyclist goes 26 minutes more towards city X, and then turns back and returns to Y. What is the maximum difference in time the pedestrian and the cyclist arrive to Y? Express the answer in minutes.

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Let $v=$ the speed of the cyclist, vp=v_p=vp​= the speed of the pedestrian.

Given 10 km/h≤v≤13 km/h,vp=5 km/h.10 km/hleq vleq 13 km/h, v_p=5 km/h.10 km/h≤v≤13 km/h,vp​=5 km/h.

Time the pedestrian arrives to Y is

tp=21 km5 km/h=4.2 h=252 mint_p=dfrac{21 km}{5 km/h}=4.2 h=252 mintp​=5 km/h21 km​=4.2 h=252 min

Time the cyclist arrives to Y is

tc=215+v(60)+26+26+215+v(60)t_c=dfrac{21}{5+v}(60)+26+26+dfrac{21}{5+v}(60)tc​=5+v21​(60)+26+26+5+v21​(60)

=25205+v+52=dfrac{2520}{5+v}+52=5+v2520​+52

The difference in time is

T=252−(25205+v+52)=1000+200v−25205+vT=252-(dfrac{2520}{5+v}+52)=dfrac{1000+200v-2520}{5+v}T=252−(5+v2520​+52)=5+v1000+200v−2520​

=200v−15205+v=dfrac{200v-1520}{5+v}=5+v200v−1520​

Tv′=200(5+v)−(200v−1520)(5+v)2=2520(5+v)2>0T_v’=dfrac{200(5+v)-(200v-1520)}{(5+v)^2}=dfrac{2520}{(5+v)^2}>0Tv′​=(5+v)2200(5+v)−(200v−1520)​=(5+v)22520​>0

Since $10km/h\le v\le 13km/h,$ then the maximum difference in time the pedestrian and the cyclist arrive to Y is

200(13)−15205+13=60(min)dfrac{200(13)-1520}{5+13}=60 (min)<span class="pstr