Question

The profit \(P\) (in dollars) from selling \(x\) units of calculus textbooks is given by \(P=-0.05 x^{2}+20 x-1000\) (a) Find the additional profit when the sales increase from 150 to 151 units. (b) Find the marginal profit when \(x=150\). (c) Compare the results of parts (a) and (b).

Short answer

The additional profit arises from an increase in sales from 150 to 151 units corresponds the approximate value of the marginal profit when the sales is at 150 units.

Step-by-step solution

Find the additional profit

To find the additional profit when sales increase from 150 to 151 units, we simply substitute these values into the profit equation. Therefore we calculate the profit for 151 units and subtract the profit for 150 units. \nProfit for 151 units: \(P_{151}=-0.05(151)^{2}+20(151)-1000\) \n Profit for 150 units: \(P_{150}=-0.05(150)^{2}+20(150)-1000\) \n Additional profit: \(\Delta P = P_{151} - P_{150}\)

Find the Marginal Profit

To find the marginal profit at \(x=150\), we have to first find the derivative of the function \(P(x)\). The derivative of \(P(x)\) represents the rate of change of profit with respect to number of units. The formula for the derivative of a quadratic equation ax² + bx + c is 2ax + b. Substituting the coefficients of the equation results in the derivative of the profit function \(P'(x)\): \n\(P'(x)=-0.1x+20\) \n Now substituting \(x=150\) into the derivative equation gives us the marginal profit when \(x=150\). \nMarginal profit, \(P'(150)=-0.1(150)+20\) \n

Compare the Results

To perform a comparison between the additional profit \( \Delta P \) and marginal profit \( P'(150) \), present both results side by side. A meaningful comparison can reveal interesting business insights such as how much the extra unit of calculus textbook brings in and whether it correlates with the marginal profit - indicative of the instantaneous rate of profit increment.

Key concepts

Marginal Profit

Marginal profit is a critical concept in economics and business. It refers to the additional profit gained from selling one more unit of a product.
Understanding marginal profit is essential for businesses as it helps in making decisions about whether or not to increase production.
To calculate the marginal profit, we look at the derivative of the profit function. The derivative essentially gives the slope of the profit function at any given point, representing the rate of change in profit. This tells us how much the profit is expected to increase when one more unit is sold.
In the problem, the marginal profit at \(x=150\) is computed by taking the derivative of our profit function \(P(x)\) and substituting \(x=150\). It's a handy way to get a sense of the profitability at this specific point in the sales of calculus textbooks.

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree two. It is characterized by the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable.
In this exercise, the profit function \(P(x) = -0.05x^2 + 20x - 1000\) is a quadratic equation. The graph of a quadratic equation is a parabola, which can open upwards or downwards depending on the sign of the leading coefficient \(a\). Here, \(a = -0.05\), which means the parabola opens downwards, indicating that there's a maximum profit point.
Quadratic equations are prevalent in real-world scenarios, particularly in cases dealing with profits, costs, and revenues, making our understanding of them crucial for interpreting business functions like the one in this problem.

Rate of Change

The rate of change is a mathematical concept that describes the change of a quantity in relation to another quantity. It is essentially the slope of the curve represented by a function.
In the context of this problem, we're interested in the rate of change of profit with respect to the number of units sold. This is represented by the derivative of the profit function.
Understanding the rate of change helps businesses and economists determine how quickly profits increase or decrease as sales volume changes, providing insights into the dynamics of the market and production efficiency.

Derivative

A derivative is a foundational concept in calculus. It measures how a function changes as its input changes, offering insights into the function's behavior.
In business mathematics, the derivative of a profit function tells us the marginal profit – the profit change for a small increase in unit sales.
For a quadratic equation, the derivative gives us a linear equation. In this case, the derivative of the profit function \(P(x) = -0.05x^2 + 20x - 1000\) is \(P'(x) = -0.1x + 20\). This new function calculates marginal profit at any point \(x\).
The process of differentiation provides valuable tools for analyzing various business functions, helping optimize production and maximizing profit by understanding how changes in sales impact overall profitability.