Question 758089
The distance walked would be (12-x) miles.
The distance rowed would be {{{sqrt(16+x^2)}}} miles.
(A) {{{highlight(T(x)=sqrt(16+x^2)/2+(12-x)/3)}}} , measured in hours.
 
(B) T'(x) = {{{(1/2)(1/2/sqrt(16+x^2))*2x-1/3=x/2/sqrt(16+x^2)-1/3}}} is zero for {{{x=c}}} so
{{{c/2sqrt(16+c^2)-1/3=0}}} --> {{{c/2sqrt(16+c^2)=1/3}}} --> {{{c/sqrt(16+c^2)=2/3}}} --> {{{c^2/(16+c^2)=4/9}}} --> {{{9c^2=4(16+c^2)}}} --> {{{9c^2=64+4c^2}}} --> {{{5c^2=64}}} --> {{{c^2=64/5}}} --> {{{c=sqrt(64/5)=8/sqrt(5)=8sqrt(5)/5}}}
The approximate value is {{{highlight(c=3.5777miles)}}}
Since you are entering an answer, the expected answer may be "3.6 miles", but I would not know what format will be accepted. Maybe "miles" must be abreviated. Maybe spaces are not accepted, and the computer wants "c=3.58mi" for an answer. Maybe an answer starting with "c=" is not accepted.
 
(C) The least travel time is {{{T(c)=sqrt(16+c^2)/2+(12-c)/3)}}}
I calculated {{{T(c)=approx5.49hours}}}
 
(D) T"(x) = {{{1/2/sqrt(16+x^2)-x^2/2/(16+x^2)^(3/2)}}}={{{8/(16+x^2)^(3/2)}}}
I calculated T"(c) = {{{approx0.052hour/mile^2}}}
I see no reason to make us calculate that. Finding that T(c)=0, while T(x)<0 for x < c and T(x)>0 for x > c proves that T(x) decreases for x < c and increases for x > c, so there is a minimum at x=c.