Problem 115 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production \(P\) with respect to 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) 3000 (B) 3500 (C) 4500 (D) 2500

Step-by-Step Solution

Answer

The new level of production is 3500 items (option B).

Step 1: Understand the problem

We need to determine the new level of production when 25 more workers are employed by the firm. We know the rate of change of production with respect to the number of workers, \( \frac{dP}{dx} = 100 - 12\sqrt{x} \), and the firm initially produces 2000 items.

Step 2: Set up the integral

To find the change in production, we integrate the rate of change function from 0 to 25, as the firm hires 25 additional workers. Calculate \( \int_{0}^{25} (100 - 12\sqrt{x}) \, dx \).

Step 3: Compute the integral

Start by integrating the function: \[ \int (100 - 12\sqrt{x}) \, dx = \int 100 \, dx - \int 12\sqrt{x} \, dx. \] The integral of \(100\) is \(100x\), and for \(12\sqrt{x}\), rewrite this as \(12x^{1/2}\). The integral is \(12 \cdot \frac{2}{3}x^{3/2} = 8x^{3/2}\).

Step 4: Evaluate the integral from 0 to 25

The definite integral from 0 to 25 for each part: \[ \int_{0}^{25} 100 \, dx = [100x]_{0}^{25} = 100 \times 25 - 100 \times 0 = 2500, \] \[ \int_{0}^{25} 12\sqrt{x} \, dx = [8x^{3/2}]_{0}^{25} = 8 \times 25^{3/2} - 8 \times 0. \] Calculate \(8 \times 25^{3/2}\): first find \(25^{3/2}\), which is \((\sqrt{25})^3 = 5^3 = 125\), so \(8 \times 125 = 1000\).

Step 5: Calculate the new production level

The change in production is \(2500 - 1000 = 1500\) additional items. Since the firm initially produces 2000 items, the new production level is \(2000 + 1500 = 3500\) items.

Key Concepts

Integration

Integration is a fundamental concept in calculus that helps us to find the accumulated quantity, such as area, volume, and in this case, the change in production. In the given exercise, integration is used to calculate how the production changes when additional workers are employed. We start with a function, specifically the rate of change of production with respect to the number of workers, \( \frac{dP}{dx} = 100 - 12\sqrt{x} \).
This function describes how production increases or decreases as more workers are hired. To find the total change in production after hiring these workers, we integrate this rate of change function over the interval from 0 to 25 workers.

• Integration effectively adds up tiny changes in production as each additional worker is hired.
• It provides us with the total change in production over this range of workers.

The process involves calculating the definite integral, which accumulates these small changes into a single value representing the total production adjustment.

Calculus

Calculus is the branch of mathematics that studies rates of change. The two main tools in calculus are differentiation and integration. Differentiation focuses on finding the rate of change, while integration is about summing up these changes.
In our problem, calculus allows us to transition smoothly from knowing how production changes with each worker (a rate) to finding out the total production change over several workers through integration.

Specifically, we often start with a derivative, \( \frac{dP}{dx} = 100 - 12\sqrt{x} \), to analyze how a small increase in x (number of workers) affects P (production).

• The derivative tells us what happens instantaneously—one worker at a time—which is crucial for making precise predictions in changing systems.
• This understanding lays the groundwork for the integration process to find cumulative change.

By applying calculus, we bridge smaller, local changes (via derivatives) to comprehensive, total shifts in production (via integrals), offering insights into what decisions, like hiring more workers, might mean for the company's output.

Definite Integral

A definite integral is a specific type of integration that calculates the total change over a set range of values. It is called "definite" because it evaluates the integral between two specific points—here, from 0 to 25 workers.
Using definite integrals, you can find not just how much something changes generally, but the exact total change over a specified interval.
This is helpful for precise planning.
In our scenario, the definite integral \( \int_{0}^{25} (100 - 12\sqrt{x}) \, dx \) helps us find how production shifts specifically by adding 25 workers to the firm.

• This integral is computed by first integrating the function components—each term individually.
• Then, we plug in the boundary values (in this case, 0 and 25) to determine the exact change over this interval.
• Finally, evaluating this definite integral shows us that production rises by 1500 items.

The definite integral thus allows us to link the theoretical rate of change to tangible results, such as how many more products the firm can produce with additional workforce.