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.
Answers
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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