Posted by Shirley on .
The equation P=240I8I^2 represents the power, P, (in watts) of a 240 volt circuit with a resistance of 8 ohms when a current of I amperes is passing through the circuit.
Find the maximum power (in watts)that can be delivered in this circuit.
I have trouble to find the maximum power,please help! THANKS A LOT!

Math 
Damon,
That is a parabola.
If you know calculus, set dP/di = 0
I will continue assuming you do not know calculus and finding the vertex of the parabola by completing the square
 8 i^2 + 240 i = P
i^2  30 i =  P/8
i^2  30 i +(30/2)^2 = P/8 + 225
(i15)^2 = P/8 + 225
vertex at i = 15 and P at 225*8 = 1800 
Math 
Shirley,
From this part on, I don't really get it:
i^2  30 i +(30/2)^2 = P/8 + 225
(i15)^2 = P/8 + 225
vertex at i = 15 and P at 225*8 = 1800
Can you please explain??? 
Math 
Damon,
I solved that quadratic equation by "completing the square".
when you have the equation in the form:
1 x^2 + b x = c
take half of b and square it
(b/2)^2 in this case (30/2)^2 = 15^2 = 225
add that to both sides
1 x^2 + b x + (b/2)^2 =  c + (b/2)^2
the left side factors
(x + b/2)(x + b/2) = c + (b/2)^2
in our case b/2 = 15 so (b/2)^2 = 225
or
(x+b/2)^2 = c + (b/2)^2
in our case
(x15)^2 = P/8 + 15^2
take sqrt of both sides remembering that  the square root is also a solution.
x+b/2 = +/ sqrt( b^2/4  4c/4)
x = ( b/2 +/ (1/2)sqrt ( b^2  4 c)
which you will recognize as the quadratic equation with a = 1
We fixed a as one by dividing by a right at the start.
The main thing is that b/2a or in our case b/2 is the axis of symmetry of the parabola. If the parabola faces up (holds water), the minimum is there. If the parabola faces down ( sheds water  our case), the maximum is there.