Let the time taken to reach the injured be #t# and the distance they land the boat from B be #x#, and we can form an equation that follows,
#t=sqrt(4^2+x^2)/6+(10-x)/10#
where #sqrt(4^2+x^2)# is the distance travelled by boat, and #10-x# is the distance travelled on foot.
Simplify,
#t=1/6(16+x^2)^(1/2)-1/10x+1#
Differentiate,
#dt/dx=1/6*1/2*2x*(16+x^2)^(-1/2)-1/10#
Simplify,
#dt/dx=(2x)/(12sqrt(16+x^2))-1/10#
Let #dt/dx=0#
#(2x)/(12sqrt(16+x^2))-1/10=0#
Add #1/10# to both sides,
#(2x)/(12sqrt(16+x^2))=1/10#
Cross multiply,
#20x=12sqrt(16+x^2)#
Square both sides,
#400x^2=144(16+x^2)#
Expand,
#400x^2=2304+144x^2#
Subtract #144x^2# from both sides,
#256x^2=2304#
Divide both sides by #256#
#x^2=9#
Square root both sides,
#x=3# ( reject #-3# as distance#>0# )
Hence, they should land #3# units from B, which is C.
