Question

1. ɪғ ᴀ ʏᴏᴜɴɢ ᴍᴀɴ ʀɪᴅᴇs ʜɪs ᴍᴏᴛᴏʀᴄʏᴄʟᴇ ᴀᴛ 25 ᴋᴍ/ʜ, ʜᴇ ʜᴀs ᴛᴏ sᴘᴇɴᴅ ₹ 2 ᴘᴇʀ ᴋᴍ ᴏɴ ᴘᴇᴛʀᴏʟ. ɪғ ʜᴇ ʀɪᴅᴇs ᴀᴛ ᴀ ғᴀsᴛᴇʀ sᴘᴇᴇᴅ ᴏғ 40 ᴋᴍ/ʜ, ᴛʜᴇ ᴘᴇᴛʀᴏʟ ᴄᴏsᴛ ɪɴᴄʀᴇᴀsᴇ ᴛᴏ ₹ 5 ᴘᴇʀ ᴋᴍ. ʜᴇ ʜᴀs ₹100 ᴛᴏ sᴘᴇɴᴅ ᴏɴ ᴘᴇᴛʀᴏʟ ᴀɴᴅ ᴡɪsʜᴇs ᴛᴏ ᴄᴏᴠᴇʀ ᴛʜᴇ ᴍᴀxɪᴍᴜᴍ ᴅɪsᴛᴀɴᴄᴇ ᴡɪᴛʜɪɴ ᴏɴᴇ ʜᴏᴜʀ. ᴇxᴘʀᴇss ᴛʜɪs ᴀs ʟ.ᴘ.ᴘ. ᴀɴᴅ ᴛʜᴇɴ sᴏʟᴠᴇ ɪᴛ ɢʀᴀᴘʜɪᴄᴀʟʟʏ.​

Answer 1

please mark brilliant

Answer 2

Given :  ʏᴏᴜɴɢ ᴍᴀɴ ʀɪᴅᴇs ʜɪs ᴍᴏᴛᴏʀᴄʏᴄʟᴇ ᴀᴛ 25 ᴋᴍ/ʜ, ʜᴇ ʜᴀs ᴛᴏ sᴘᴇɴᴅ ₹ 2 ᴘᴇʀ ᴋᴍ ᴏɴ ᴘᴇᴛʀᴏʟ. ɪғ ʜᴇ ʀɪᴅᴇs ᴀᴛ ᴀ ғᴀsᴛᴇʀ sᴘᴇᴇᴅ ᴏғ 40 ᴋᴍ/ʜ, ᴛʜᴇ ᴘᴇᴛʀᴏʟ ᴄᴏsᴛ ɪɴᴄʀᴇᴀsᴇ ᴛᴏ ₹ 5 ᴘᴇʀ ᴋᴍ

To find : ᴍᴀxɪᴍᴜᴍ ᴅɪsᴛᴀɴᴄᴇ ᴡɪᴛʜɪɴ ᴏɴᴇ ʜᴏᴜʀ

Solution:

25 km/hr    - T₁  hrs

40 km/h   - T₂ hrs

T₁+ T₂ ≤ 1

Cost  =( 25T₁ ) 2  +  (40T₂ )5

C = 50T₁ + 200T₂

50T₁ + 200T₂   ≤ 100

=>  T₁ + 4T₂   ≤  2

Distance = 25T₁  + 40T₂  

From the graph we get boundary points

T₁ ,  T₂   as ( 0 , 1/2)  , ( 2/3 , 1/3) , ( 1 , 0)

Let check Distance in each case

T₁ ,  T₂    Cost       Distance

0     1/2    100        20

2/3   1/3    100       30

1        0      50        25

Hence maximum Distance covered  = 30 km

when at 25 km/hr for  2/3 hrs ( 40 mins)  & at 40 km/hr  for 1/3 hrs ( 20 mins)

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