Question

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

Answer 1

Given:

The power generated by an electrical circuit (in watts) as function of its current x (in amperes) is modeled by:

$P(x)=-12x^2+120x$

To find:

The current that will produce the maximum power.

Solution:

We have,

$P(x)=-12x^2+120x$

Here, leading coefficient is negative. So, it is a downward parabola.

Vertex of a downward parabola is the point of maxima.

If a parabola is $f(x)=ax^2+bx+c$, then

$Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)$

In the given function, a=-12 and b=120. So,

$-\dfrac{b}{2a}=-\dfrac{120}{2(-12)}$

$-\dfrac{b}{2a}=-\dfrac{120}{-24}$

$-\dfrac{b}{2a}=5$

Putting x=5 in the given function, we get

$P(5)=-12(5)^2+120(5)$

$P(5)=-12(25)+600$

$P(5)=-300+600$

$P(5)=300$

Therefore, 5 watt current will produce the maximum power of 300 amperes.