Question 1187611

The first derivative is y'=2x-(18/x)
set that equal to 0 and 2x=(18/x) and 2x^2=18 and x^2=9 and x=+/-3.  Only +3 is in the interval.
So the critical values are x=1, 3, 6
f(1)=1, since ln 1=0
f(3)=9-18 ln 3=-10.78 This is the minimum
f(6)=36-18ln 6=3.748 This is the maximum

Graph of the function is
{{{graph(300,300,-1,7,-20,10,x^2-18ln(x))}}}