Problem 43 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P w.r.t. additional number of workers \(x\) is given by \(\frac{d P}{d x}=100-12 \sqrt{x}\). If the firm employs 25 more workers, then the new level of production of items is (a) 2500 (b) 3000 (c) 3500 (d) 4500

Step-by-Step Solution

Answer

The new level of production is 3500 items.

Step 1: Understand the Given Information

You are given that the original production of items is 2000. The rate of change of production with respect to additional workers is expressed as \( \frac{dP}{dx} = 100 - 12\sqrt{x} \). This describes how the production changes as more workers are employed.

Step 2: Set Up the Integral for Total Change in Production

To find the total change in production when 25 additional workers are employed, you integrate the rate \( \frac{dP}{dx} \) over the interval from \( x = 0 \) to \( x = 25 \), since no extra workers were initially employed. The integral to evaluate is \( \int_0^{25} (100 - 12\sqrt{x}) \, dx \).

Step 3: Integrate the Expression

Break down the integral as follows: \[\int_0^{25} (100 - 12\sqrt{x}) \, dx = \int_0^{25} 100 \, dx - \int_0^{25} 12\sqrt{x} \, dx\]The first part is \( 100 \int_0^{25} dx \) and the second is \( 12 \int_0^{25} x^{1/2} dx \).

Step 4: Integrate Individual Components

Evaluate both parts separately:- \( \int_0^{25} 100 \, dx = 100[x]_0^{25} = 100(25 - 0) = 2500 \).- \( \int_0^{25} 12\sqrt{x} \, dx = 12\left[ \frac{2}{3}x^{3/2} \right]_0^{25} = 12 \times \frac{2}{3} \times (25^{3/2} - 0^{3/2}) \). - Compute \( 25^{3/2} = (5^2)^{3/2} = 5^3 = 125 \).

Step 5: Simplify and Calculate

Calculate the result:- For the \( 12\left[ \frac{2}{3}x^{3/2}\right]_0^{25} \) integral: \( 12 \times \frac{2}{3} \times 125 = 1000 \).- Thus, the total change is \( 2500 - 1000 = 1500 \).

Step 6: Determine the New Production Level

Add the change in production, which is 1500, to the original production, which was 2000. This gives: - New production level = 2000 + 1500 = 3500.

Key Concepts

Integration in Calculus

Integration in calculus is a fundamental concept that involves finding the overall change in a quantity over an interval. Essentially, integration is the reverse process of differentiation. In the context of this exercise, integration helps us determine how the production levels of a company change as more workers are hired.

When you have a rate of change given by a derivative, such as \(\frac{dP}{dx} = 100 - 12\sqrt{x}\), integration allows you to accumulate this rate over a specified range. This range is from \(x=0\) to \(x=25\), representing the number of additional workers. The goal of integrating \(\int_0^{25} (100 - 12\sqrt{x}) \, dx\) is to find the total production increase due to employing 25 more workers.

Here, the function \(100 - 12\sqrt{x}\) is taken and integrated. This means essentially 'adding up' the production changes over every small increase in workers, from 0 to 25. This gives a complete picture of how much total production increases. Integration helps us calculate this precise figure by using calculus techniques.

Differential Equations

Differential equations play a critical role in understanding relationships between varying quantities. They involve rates of change and can be solved to predict future behavior of dynamic systems. In our exercise, knowing the rate of change of production \(\frac{dP}{dx}\) is the differential equation that governs how production levels react to the addition of workers.

A differential equation in this context represents a situation where the change in production depends on the number of workers. The equation \(\frac{dP}{dx} = 100 - 12\sqrt{x}\) means that as more workers are hired, the impact on production is initially high but decreases as \(x\) (the number of additional workers) increases.

Solving this differential equation involves integrating to revert the derivative and find the total change in production, a solution that provides meaningful insights into the company's operational strategies. Through integration, we learn the cumulative effect on production when adding 25 workers.

Applications of Calculus

Calculus has numerous practical applications, especially in economics and business, where understanding changes in production or costs is crucial for decision-making. This exercise is a classic example of calculus in action, applied to understand how hiring decisions affect production levels.

Using calculus, businesses like this firm can forecast how varying the number of workers impacts productivity. By understanding the function \(\frac{dP}{dx} = 100 - 12\sqrt{x}\), decision-makers can predict future production outputs based on different employment levels. The integration of this function gives the net increase in production, guiding informed strategic adjustments.

Ultimately, calculus provides a scientific framework for assessing changes, optimizing resources, and planning for future demand. With these models, firms can fine-tune their operations to maximize efficiency and productivity, demonstrating the power of mathematical analysis in everyday business practice.