To determine the number of units to be produced by the firm to maximize its profit, we need to establish the relationship between the firm's costs, its revenue, and the resulting profit. Profit is the primary goal for many firms, and it is achieved by finding the ideal level of production where the difference between total revenue and total cost is the largest.

Profit Maximization Principles

A fundamental principle in economics states that a firm maximizes its profit when its marginal revenue (MR) equals its marginal cost (MC). From a calculus perspective, this occurs when the first derivative of the profit function with respect to the quantity produced is zero, and the second derivative is negative, confirming a maximum point.

• Cost Function: This function, denoted as \(C(q)\), describes the total expenses incurred for producing \(q\) units of a product.
• Revenue Function: The total revenue \(R(q)\) is generated by selling \(q\) units at a given market price per unit.
• Profit Function: The profit function \(\Pi(q)\) is derived by subtracting the total cost from the total revenue.

Cost and Revenue Functions

Let's define the specific cost and revenue functions based on the information provided in the question:

1. Cost Function (\(C(q)\)):

   The question states that the cost function for a product is given by \(5q^2\), where \(q\) represents the amount of production.

   $$C(q) = 5q^2$$

2. Revenue Function (\(R(q)\)):

   The firm can sell the product at a market price of Rs. 50 per unit. To find the total revenue for \(q\) units, we multiply the price per unit by the quantity produced:

   $$R(q) = \text{Price per unit} \times \text{Quantity}$$

   $$R(q) = 50 \times q$$

   $$R(q) = 50q$$

Profit Function Derivation

The profit function, \(\Pi(q)\), is defined as total revenue minus total cost. We substitute the expressions for \(R(q)\) and \(C(q)\) that we derived:

$$\Pi(q) = R(q) - C(q)$$

$$\Pi(q) = 50q - 5q^2$$

Optimal Production Calculation

To find the number of units (\(q\)) that maximizes the firm's profit, we use calculus. We need to find the production level where the rate of change of profit with respect to quantity is zero. This is done by taking the first derivative of the profit function and setting it equal to zero.

1. First Derivative of the Profit Function:

   Differentiate \(\Pi(q)\) with respect to \(q\):

   $$\frac{d\Pi}{dq} = \frac{d}{dq}(50q - 5q^2)$$

   $$\frac{d\Pi}{dq} = 50 - 10q$$

2. Set the First Derivative to Zero:

   To find the quantity that maximizes profit, we set the first derivative to zero:

   $$50 - 10q = 0$$

3. Solve for \(q\):

   Now, we solve the equation for \(q\):

   $$10q = 50$$

   $$q = \frac{50}{10}$$

   $$q = 5$$

4. Second Derivative Test (for verification):

   To confirm that \(q=5\) indeed maximizes profit, we check the sign of the second derivative of the profit function. A negative second derivative indicates a maximum point.

   $$\frac{d^2\Pi}{dq^2} = \frac{d}{dq}(50 - 10q)$$

   $$\frac{d^2\Pi}{dq^2} = -10$$

   Since the second derivative is \(-10\), which is less than zero (\(-10 < 0\)), it confirms that the profit is maximized when the firm produces 5 units.

Therefore, the number of units to be produced by the firm such that the profit is maximized is 5.