Let company increases the annual subscription by \(Rs\ x\).
So, x subscribers will discontinue the service.
Total revenue of company after the increment
\(R(x) =(500−x)(300+x)\)
\(R(x) =1500000+500x−300x−x^2\)
\(R(x) = −x^2+200x+150000\)
Differentiate both sides w.r.t, x
\(R^′(x)=−2x+200\)
Now, \(R^′(x)=0\)
\(2x=200\)
\(x=100\)
∴ \(R^{''}(x)=−2<0\)
R(x) is maximum when \(x = 100\).
Therefore, the company should increase the subscription fee by \(Rs.\ 100\), so that it has maximum revenue.
So, the correct option is (A): \(100\)
Area of region enclosed by curve y=x3 and its tangent at (–1,–1)
The minimum of \(f(x)=\sqrt{(10-x^2)}\) in the interval \([-3,2]\) is
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
Read More: Application of Derivatives