Answer:
Option A
Step-by-step explanation:
A hunter travels to his cabinet located at 2 Miles North and 8 Miles West.
Let "d" be the distance traveled before reaching the rocky ground at point say "P". Let the cabin be at "C"
The distance between C and P is given by
[tex]\sqrt{(2^2+(8-d)^2}\\[/tex]
Let "t" be the time taken to travel distance CP.
[tex]T = \frac{d}{5} + \frac{\sqrt{2^2+(8-d)^2} }{3}[/tex]
Minimum time will be when "t" is equal to zero
Differentiating the above equation wrt "d", we get -
[tex]= \frac{1}{5} + \frac{1}{3} (\frac{-2(8-d)}{2\sqrt{2^2+(8-d)^2} } )= 0\\\frac{1}{5} = \frac{1}{3} \frac{(8-d)}{\sqrt{4+(8-d)^2} } \\3\sqrt{4+64+d^2-16d} = 40-5d\\[/tex]
On squaring both the sides , we get -
[tex]3\sqrt{4+64+d^2-16d} = 40-5d\\9(68+d^2-16d)= 1600+25d^2 -400d\\612+9d^2-144d=1600+25d^2 -400d\\16d^2-256d+988=0\\d_1 = 9.5\\d_2 = 6.5[/tex]
"d" cannot be greater than 8 miles.
Hence, the value of "d" is equal to [tex]6.5[/tex]
Option A