The equation P=240I-8I^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!

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

(i-15)^2 = -P/8 + 225

vertex at i = 15 and P at 225*8 = 1800

From this part on, I don't really get it:

i^2 - 30 i +(30/2)^2 = -P/8 + 225

(i-15)^2 = -P/8 + 225

vertex at i = 15 and P at 225*8 = 1800

Can you please explain???

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
(x-15)^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.

To find the maximum power that can be delivered in this circuit, we need to determine the value of current (I) that maximizes the power equation P = 240I - 8I^2. This can be done by finding the vertex of the parabolic equation.

In general, the vertex of a parabola with the equation y = ax^2 + bx + c is given by the x-coordinate x = -b/(2a).

In this case, the equation for power is P = 240I - 8I^2, which means the coefficient of I^2 is -8, and the coefficient of I is 240.

To find the value of current that maximizes power, we need to find the x-coordinate of the vertex:

x = -b/(2a) = -240/(2(-8)) = -240/(-16) = 15

So, the value of current (I) at which the maximum power is achieved is 15 amperes.

To find the maximum power, substitute the value of I = 15 back into the power equation:

P = 240I - 8I^2
P = 240(15) - 8(15)^2
P = 3600 - 8(225)
P = 3600 - 1800
P = 1800

Therefore, the maximum power that can be delivered in this circuit is 1800 watts.