Question 284312
Find the line connecting (0,3) and (2,0).
{{{m=(0-3)/(2-0)=-3/2}}}
{{{y=mx+b=-(3/2)x+b}}}
When x=0, y=3,
{{{y=-(3/2)(0)+b=3}}}
{{{b=3}}}
So then
{{{y=-(3/2)x+3}}}
Find {{{y^2}}}
{{{y^2=(-(3/2)x+3)^2=(9/4)x^2-9x+9}}}
{{{2y^2=(9/2)x^2-18x+18}}}
Substitute into P(x,y),
{{{P(x,y)=3x^2+2y^2=3x^2+(9/2)x^2-18x+18}}}
{{{P(x)=(15/2)x^2-18x+18}}}
Now P is only a function of x. 
Differentiate wrt x and set the derivative equal to zero to find the minimum.
{{{dP/dx=(15/2)(2x)-18=15x-18=0}}}
{{{15x=18}}}
{{{x=18/15=6/5}}}
Now going back to the line equation,
{{{y=-(3/2)x+3}}}
{{{y=-(3/2)(6/5)+3=-9/5+15/5=6/5}}}
The minimum occurs at (6/5,6/5).