Question 907484
Joe's kayak is on the lake at point {{{K}}} (for KAYAK).
The nearest point on the shore is {{{N}}} (for NEAREST),
and Joe's car is next to the shoreline at point {{{C}}} (for CAR).
Joe will land the kayak at shoreline point {{{L}}} (for LANDING),
which is somewhere between {{{N}}} and {{{C}}} ,
at a distance {{{x}}} {{{miles}}} from point {{{N}}} ,
at a distance {{{h}}} (for hypotenuse) from point {{{K}}},
and at a distance {{{5-x}}} from point  {{{C}}} .


{{{drawing(400,200,-1,7,-1,3,
line(-1,0,7,0),circle(5,0,0.05),
triangle(0,0,0,2,1.79,0),
rectangle(0,0,0.2,0.2),
locate(5.5,0,shoreline),
locate(-0.1,0,N),locate(-0.1,2.3,K),
locate(1.7,0,L),locate(4.9,-0.1,C),
locate(0.1,1.1,2),locate(0.8,0.3,x),
locate(0.8,1,h),locate(3,0.3,5-x),
arrow(2.8,0.15,1.8,0.15),arrow(3.8,0.15,5,0.15)
)}}}
The Pythagorean theorem tells us that
{{{h=sqrt(x^2+2^2)=sqrt(x^2+4)}}} .
The time (in hours) for Joe to cover that distance kayaking at 2 MPH is
{{{sqrt(x^2+4)/2}}} .
The time (in hours) for Joe to cover the distance {{{5-x}}} walkking at 3 MPH is
{{{(5-x)/3}}}
The total time (in hours for Joe to go from point {{{K}}} to point {{{C}}} is
{{{t(x)=sqrt(x^2+4)/2+(5-x)/3}}}
( {{{t}}} as a function of {{{x}}} ).
 
That much you would reason, regardless of your math class level and instructor preferences.
From this point on, what your instructor wants depends.
If you are being pushed to use and abuse a graphing calculator, you may be expected to graph the function {{{t(x)}}} and find its minimum.
If you are studying calculus, you may be expected to calculate the derivative of  {{{t(x)}}} to find the point where the derivative is zero and {{{t(x)}}} is minimum:
{{{dt/dx=(1/2)(1/2sqrt(x^2+4))(2x)+(1/3)(-1)=x/2sqrt(x^2+4)-1/3=(3x-2sqrt(x^2+4))/6sqrt(x^2+4)}}}
{{{dt/dx=0}}} ---> {{{3x-2sqrt(x^2+4)=0}}} ---> {{{3x=2sqrt(x^2+4)}}} ---> {{{9x=4(x^2+4)}}} ---> {{{9x=4x^2+16}}} ---> {{{5x^2=16}}} ---> {{{x^2=16/5}}} ---> {{{x=sqrt(16/5)}} ---> {{{x=4/sqrt(5)=4sqrt(5)/5}}}
So {{{highlight(x=about 1.79)}}} .
(Joe should aim for landing {{{highlight(1.79)}}} miles from point {{{N}}}.
Now that we have found {{{x}}} for Joe,
he has all the side lengths for right triangles {{{NKL}}} (pictured) and {{{NKC}}} (imagine it),
so I would let Joe figure out the angles {{{NKL}}} , {{{NKC}}} and {{{LKC}}}
that he can use to get going in the right direction.