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

The current that produces the maximum power is 5 amperes.

Step-by-step explanation:

To find the maximum power, we need to analyze the extrema (maximum and minimum) of the function P(x).

The derivative of P(x) is -24x + 120. Setting this equal to 0 and solving for x gives us the critical point: x = 5.

Since the second derivative of P(x) is -24 (negative), we know the critical point is a maximum. Therefore, the current of 5 amperes generates the maximum power.

Alternatively, we can notice that P(x) is a quadratic function with a negative leading coefficient, meaning it opens downwards and has a single maximum point at its vertex. The vertex's x-coordinate coincides with the critical point, again confirming the current of 5 amperes yields the maximum power.

Answer 2

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((-b)/(2a),f((-b)/(2a))\right)`

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

`-(b)/(2a)=-(120)/(2(-12))`

`-(b)/(2a)=-(120)/(-24)`

`-(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.