So now you should be able to write out T(x), which needs to be in units of time, as:
1. T(x) = [ √(2^2+x^2) ] / (3 miles/hr) + ( 6 - x ) / (4 miles/hr)
2. T(x) = (√[x^2+4])/3 + 6/4 - x/4
If you look at it, the distances are in the numerators and the speeds are in the denominators. Distance divided by speed gives time. So the above adds up two times -- the time by water plus the time by land.
To "optimize" this, you need to find the derivative and solve it with respect to x and set that equal to 0 (which means either a minimum or maximum) and then find x from that. I assume that is about where your class is at, right now. Anyway, so take the derivative:
3. d( T(x) ) = d( (√[x^2+4])/3 + 6/4 - x/4 )
4. d( T(x) ) = d( (√[x^2+4])/3 ) + d( 6/4 ) - d( x/4 )
5. d( T(x) ) = d( (√[x^2+4])/3 ) - d( x/4 )
6. d( T(x) ) = (1/3)*d( √[x^2+4] ) - 1/4 dx
7. d( T(x) ) = (1/3)*( x/√[x^2+4] ) dx - 1/4 dx
8. d( T(x) )/dx = x/(3√[x^2+4]) - 1/4 = 0
That solves out as:
9. x = 6/√7 miles
That should minimize the time. Now you need to plug that value back into equation (2) above to figure out the least travel time. 1.941 hours, I think. Do a basic verification by assuming she takes the boat straight to shore, at 2/3 hours, and then walks the 6 miles, at 6/4 hours, and get a total time of 2.167 hours. Then assume she goes entirely by water, for 6.325 miles, dividing that by 3 miles/hr, and also get over 2 hours, too. So perhaps the answer is right. At least it is smaller than either of the two extremes.
But we are talking about only 10-15 minutes difference vs the 2 hour trip, so I think she should do what her mood suggests at the time, unless she is in a hurry.