Answer:
The weekly revenue is maximum at x=1.67.
Explanation:
The given function is
[tex]f(x)=1000x -300x^2[/tex] .... (1)
where, f(x) is the total revenue at price x.
We need to find the price x at which the weekly revenue is maximum.
The leading coefficient of the given function is -300, which is a negative number. So, it is a downward parabola and vertex of a downward parabola is the the point of maxima.
If a parabola is defined as
[tex]g(x)=ax^2+bx+c[/tex] ... (2)
then the vertex of the function is
[tex](-\frac{b}{2a},g(-\frac{b}{2a}))[/tex]
From (1) and (2) it is clear that
[tex]a=-300,b=1000, c=0[/tex]
The given function is maximum at
[tex]-\frac{b}{2a}=-\frac{1000}{2(-300)}[/tex]
[tex]-\frac{b}{2a}=\frac{10}{6}[/tex]
[tex]-\frac{b}{2a}=1.66667[/tex]
[tex]-\frac{b}{2a}\approx 1.67[/tex]
Therefore the weekly revenue is maximum at x=1.67.
