There are two ways to do this kind of problem. One is to use calculus, which leaves you with the equation
2I - 30 = 0, leading right away to I = 15. The other way, which does not require calculus, is to use the method of "completing the square". That is what was done in the example you posted. When the equation
P = 240I-8I^2 is rewritten
I^2 - 30 I +(30/2)^2 = -P/8 + 225
= (I-15)^2 , or
P/8 = 225 -(I - 15)^2
you see right away that the maximum value of P is obtained when I = 15.
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 had posted this question before, and someone gave me this solution:
- 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 anyone please explain it to me???THANKS A LOT!
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