Step-by-Step Solution

Topic: Application of Derivatives (Maxima & Minima)

(I) REVENUE FUNCTION R(x)

Given Data:

• Current Readers = 10,000
• Current Fee = ₹ 4,000
• Loss Rate: 10 readers for every ₹ 5 increase.

Step 1: Determine Variables

Let the total increase in subscription be ₹ x.

Number of ₹ 5 increments = x/5.

Number of readers lost = 10 * (x/5) = 2x.

Step 2: Formulate Function

New Subscription Fee = 4000 + x

New Number of Readers = 10000 – 2x

Revenue R(x) = (New Fee) * (New Readers)

R(x) = (4000 + x)(10000 – 2x)

R(x) = 40000000 – 8000x + 10000x – 2x²

(II) DERIVATIVE OF R(x)

We need to find the derivative of R(x) with respect to x.

R(x) = -2x² + 2000x + 40000000

Using the power rule:

d/dx (R(x)) = d/dx (-2x²) + d/dx (2000x) + d/dx (40000000)

d/dx (R(x)) = -4x + 2000

(III)(A) OPTIMAL INCREASE FOR MAX EARNINGS

To find the maximum earnings, set the first derivative to zero (Critical Point).

R'(x) = 0

2000 – 4x = 0

4x = 2000

x = 500

Verification (Second Derivative Test):

R”(x) = -4 (Negative)

Since the second derivative is negative, x = 500 corresponds to a Maximum.

(III)(B) MAXIMUM VALUE OF R(x)

Substitute the optimal increase x = 500 back into the revenue function R(x).

R(500) = (4000 + 500)(10000 – 2(500))

R(500) = (4500)(10000 – 1000)

R(500) = (4500)(9000)

Calculation:

45 * 9 = 405

Append five zeros: 40,500,000