SOLUTION: the electronic potential at a point (x;y) on the same line segment extending from (0;3) to (2;0) is given by {{{P=3x^2+2y^2}}} at what point on this line segment is the potentail a

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: the electronic potential at a point (x;y) on the same line segment extending from (0;3) to (2;0) is given by {{{P=3x^2+2y^2}}} at what point on this line segment is the potentail a      Log On


   



Question 284312: the electronic potential at a point (x;y) on the same line segment extending from (0;3) to (2;0) is given by P=3x%5E2%2B2y%5E2 at what point on this line segment is the potentail a minimum?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Find the line connecting (0,3) and (2,0).
m=%280-3%29%2F%282-0%29=-3%2F2
y=mx%2Bb=-%283%2F2%29x%2Bb
When x=0, y=3,
y=-%283%2F2%29%280%29%2Bb=3
b=3
So then
y=-%283%2F2%29x%2B3
Find y%5E2
y%5E2=%28-%283%2F2%29x%2B3%29%5E2=%289%2F4%29x%5E2-9x%2B9
2y%5E2=%289%2F2%29x%5E2-18x%2B18
Substitute into P(x,y),
P%28x%2Cy%29=3x%5E2%2B2y%5E2=3x%5E2%2B%289%2F2%29x%5E2-18x%2B18
P%28x%29=%2815%2F2%29x%5E2-18x%2B18
Now P is only a function of x.
Differentiate wrt x and set the derivative equal to zero to find the minimum.
dP%2Fdx=%2815%2F2%29%282x%29-18=15x-18=0
15x=18
x=18%2F15=6%2F5
Now going back to the line equation,
y=-%283%2F2%29x%2B3
y=-%283%2F2%29%286%2F5%29%2B3=-9%2F5%2B15%2F5=6%2F5
The minimum occurs at (6/5,6/5).