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.