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!
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.
Certainly! Let me explain it step by step for you.
The given equation is P = 240I - 8I^2.
To find the maximum power that can be delivered in this circuit, we need to find the vertex of the quadratic equation.
Step 1: Rewrite the equation in the form of a quadratic equation of the form ax^2 + bx + c = 0.
Rewriting the given equation, we have:
P = -8I^2 + 240I
Step 2: Rearrange the equation to set it equal to zero.
0 = -8I^2 + 240I - P
Step 3: Now, let's complete the square to find the vertex of the quadratic equation.
In this step, we are adding and subtracting a constant term (c) to both sides of the equation, where c = (b/2)^2.
To find the constant term, take the coefficient of the linear term (in this case, 240I) which is b, divide it by 2, and then square it.
c = (240/2)^2 = 30^2 = 900
Add and subtract 900 on the right-hand side:
0 = -8I^2 + 240I + 900 - 900 - P
0 = -8I^2 + 240I + 900 - P
= (-8I^2 + 240I + 900) - (900 + P)
= [-8(I^2 - 30I + 900)] - (900 + P)
Step 4: Factor the perfect square trinomial (I^2 - 30I + 900).
The perfect square trinomial factors into (I - 15)^2.
0 = [-8(I - 15)^2] - (900 + P)
Step 5: Isolate the vertex form of the equation by moving the remaining terms to the other side.
8(I - 15)^2 = - (900 + P)
Now, divide throughout by 8 to simplify:
(I - 15)^2 = - (900 + P)/8
Step 6: We have now arrived at the vertex form of the equation: (x - h)^2 = k.
Comparing the equation with the vertex form, we can see that the vertex is located at (h, k) = (15, -(900 + P)/8).
Since the vertex represents the maximum (or minimum) point of a quadratic equation, we can conclude that the maximum power that can be delivered in this circuit is attained when the current, I, is 15 amperes.
Finally, to find the maximum power (P), substitute the value of I (15) back into the original equation:
P = 240I - 8I^2
P = 240(15) - 8(15^2)
P = 3600 - 1800
P = 1800 watts
So, the maximum power that can be delivered in this circuit is 1800 watts when the current is 15 amperes.