Given:
The power generated by an electrical circuit (in watts) as function of its current x (in amperes) is modeled by:
[tex]P(x)=-12x^2+120x[/tex]
To find:
The current that will produce the maximum power.
Solution:
We have,
[tex]P(x)=-12x^2+120x[/tex]
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 [tex]f(x)=ax^2+bx+c[/tex], then
[tex]Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)[/tex]
In the given function, a=-12 and b=120. So,
[tex]-\dfrac{b}{2a}=-\dfrac{120}{2(-12)}[/tex]
[tex]-\dfrac{b}{2a}=-\dfrac{120}{-24}[/tex]
[tex]-\dfrac{b}{2a}=5[/tex]
Putting x=5 in the given function, we get
[tex]P(5)=-12(5)^2+120(5)[/tex]
[tex]P(5)=-12(25)+600[/tex]
[tex]P(5)=-300+600[/tex]
[tex]P(5)=300[/tex]
Therefore, 5 watt current will produce the maximum power of 300 amperes.
