Question

The power generated by an electrical circuit (in watts) as a function of its current x (in amperes) is modeled by: P(x)=-12x^2+120x What current will produce the maximum power?

Answer 1

Answer:

A current of 5 amperes will produce the maximum power.

Step-by-step explanation:

Let be $p(x) = -12\cdot x^{2}+120\cdot x$, where $p(x)$ is measured in watts and $x$ in amperes. At first we must obtain the first and second derivatives of the function to determine the current associated with maximum power. That is:

First derivative

$p'(x) = -24\cdot x + 120$

Second derivative

$p''(x) = -24\cdot x$

Now, we equalize the first derivative to zero and solve it afterwards: (First Derivative Test)

$-24\cdot x + 120 = 0$

$x = 5\,A$

The only critical point is $x = 5\,A$.

As next step we need to assure that critical point leads to an absolute maximum by evaluating the critical point found above in the second derivative: (Second Derivative Test)

$p(5)'' = -24\cdot (5)$

$p''(5) = -120$

Which indicates that critical point leads to an absolute maximum.

A current of 5 amperes will produce the maximum power.