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April 20, 2014

April 20, 2014

Posted by **Shirley** on Saturday, February 23, 2008 at 7:02pm.

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**, Saturday, February 23, 2008 at 7:12pmThat 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

- Math -
**Shirley**, Saturday, February 23, 2008 at 8:03pmFrom 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???

- Math -
- Math -
**Damon**, Saturday, February 23, 2008 at 9:29pmI 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.

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