Question

How many subscribers will discontinue after an increase of Rs x in annual fee ?

Answer 1

Concept:

• The rate of decrease in the number of subscribers is constant with respect to the increase in fee.
• If \( k \) subscribers discontinue for every Rs 1 increase, then for an increase of Rs \( x \), \( k \cdot x \) subscribers will discontinue.

Step 1: Identify the rate of discontinuation
According to the problem, for every Rs 1 increase in the annual subscription fee, the number of subscribers that discontinue is 10.
\[ \text{Rate of discontinuation} = 10 \text{ subscribers / Rs 1 increase} \]

Step 2: Calculate the total discontinued subscribers for Rs x increase
If the fee is increased by Rs \( x \), the total number of subscribers who will discontinue is given by the product of the rate and the amount of increase.
\[ \text{Number of discontinued subscribers} = 10 \times x = 10x \]

Answer 2

Concept:
• Total Revenue (\( R \)) = (Number of Subscribers) \(\times\) (Subscription Fee per subscriber).
• New Fee = Original Fee + Increase (\( x \)).
• New Number of Subscribers = Original Number - Discontinued Subscribers.

Step 1: Determine the new fee per subscriber
Original annual subscription fee = Rs 300. Increase in fee = Rs \( x \). \[ \text{New Fee} = (300 + x) \]

Step 2: Determine the new number of subscribers
Original number of subscribers = 5000. Number of subscribers who discontinue = \( 10x \). \[ \text{Remaining Subscribers} = (5000 - 10x) \]

Step 3: Formulate the revenue function \( R(x) \)
\[ R(x) = \text{New Fee} \times \text{Remaining Subscribers} \] \[ R(x) = (300 + x)(5000 - 10x) \] \[ R(x) = 1500000 - 3000x + 5000x - 10x^2 \] \[ R(x) = -10x^2 + 2000x + 1500000 \]

Answer 3

Concept:
• To maximize a function, find its first derivative and set it to zero (\( R'(x) = 0 \)) to find critical points.
• Use the second derivative test (\( R''(x) < 0 \)) to confirm a local maximum.

Step 1: Find the first derivative of \( R(x) \)
From the previous part: \[ R(x) = -10x^2 + 2000x + 1500000 \] Differentiating with respect to \( x \): \[ R'(x) = \frac{d}{dx}(-10x^2 + 2000x + 1500000) \] \[ R'(x) = -20x + 2000 \]

Step 2: Solve for the critical point
Set \( R'(x) = 0 \) to find the value of \( x \): \[ -20x + 2000 = 0 \] \[ 20x = 2000 \] \[ x = \frac{2000}{20} = 100 \]

Step 3: Verify the maximum using the second derivative
Differentiating \( R'(x) \) again: \[ R''(x) = \frac{d}{dx}(-20x + 2000) = -20 \] Since \( R''(100) = -20 \), which is less than 0, the function \( R(x) \) is maximized at \( x = 100 \).

Answer 4

Concept:

• A function is increasing on an interval if its first derivative is positive (\( R'(x) > 0 \)).
• A function is decreasing on an interval if its first derivative is negative (\( R'(x) < 0 \)).

Step 1: Identify the first derivative and the boundary of the domain
We found \( R'(x) = -20x + 2000 \).
The total subscribers cannot be negative: \( 5000 - 10x \geq 0 \implies x \leq 500 \).
Thus, the practical domain for the increase \( x \) is \( [0, 500] \). The question specifies a broader range (0, 5000), but we analyze based on the derivative sign.

Step 2: Find the interval where \( R(x) \) is increasing
Set \( R'(x) > 0 \):
\[ -20x + 2000 > 0 \] \[ 2000 > 20x \implies x < 100 \] So, \( R(x) \) is increasing in the interval \( (0, 100) \).

Step 3: Find the interval where \( R(x) \) is decreasing
Set \( R'(x) < 0 \):
\[ -20x + 2000 < 0 \] \[ 2000 100 \]
So, \( R(x) \) is decreasing in the interval \( (100, 5000) \). Note: realistically, it stops at \( x=500 \).