Problem 72 - Precalculus with Limits

Question

PROFIT The profit \(P\) (in hundreds of dollars) that a company makes depends on the amount \(x\) (in hundreds of dollars) the company spends on advertising. The profit function is given by \(P(x) = 200 + 30x - 0.5x^2\). Using your knowledge of the slopes of tangent lines, show that the profit is increasing on the interval \([0, 20]\) and decreasing on the interval \([40, 60]\).

Step-by-Step Solution

Answer

Using the derivative of the given profit function, it can be inferred that the profit is increasing on the interval \([0, 20]\) and decreasing on the interval \([40,60]\).

Step 1: Finding the Derivative

The derivative of the profit function \(P(x)\) is found using the power rule of derivatives. The power rule states that the derivative of \(x^n\) (where n is any real number) is \(nx^{n-1}\). Applying this rule, the derivative of \(P(x) = 200 + 30x - 0.5x^2\) is \(P'(x) = 30 - x\).

Step 2: Find where the Derivative is Positive and where it is Negative

The derivative \(P'(x) = 30 - x\) is positive for \(x < 30\) and negative for \(x > 30\). This indicates that the function \(P(x)\) is increasing in the interval \(x < 30\) and is decreasing in the interval \(x > 30\).

Step 3: Compare with the Intervals given in the Problem

The intervals given in the problem are \([0, 20]\) and \([40, 60]\). Comparing these with the intervals determined in step 2: on \([0, 20]\), \(P'(x)\) is positive and hence \(P(x)\) is increasing. On \([40, 60]\), \(P'(x)\) is negative and hence \(P(x)\) is decreasing.

Key Concepts

Power Rule

The power rule is a fundamental concept in calculus for finding the derivative of power functions. It's a straightforward method that helps us determine how a function changes at any point. Given a function of the form \(x^n\), the power rule states that its derivative is \(nx^{n-1}\). This means we multiply the exponent \(n\) by the coefficient of \(x\) and then subtract 1 from the exponent.
For the profit function \(P(x) = 200 + 30x - 0.5x^2\), we apply the power rule to each term.

• The constant term 200 has a derivative of 0 since constants do not change.
• The derivative of the linear term \(30x\) becomes 30 because the power of \(x\) is 1 and thus \(1\cdot30x^{1-1} = 30x^0 = 30\).
• For the quadratic term \(-0.5x^2\), applying the power rule gives \(-1.0x^{1}\), simplifying to \(-x\).

Hence, the derivative \(P'(x)\) is \(30 - x\). This derivative provides valuable insights into the behavior of the profit function, such as points where the profit is increasing or decreasing.

Increasing and Decreasing Functions

In calculus, a function is said to be increasing on an interval if its derivative is positive throughout that interval. Conversely, if the derivative is negative, the function is decreasing. The signs of the derivative tell us whether the slope of the tangent at any point on the curve is going upwards or downwards.
For the profit function \(P'(x) = 30 - x\), we can determine where the function is increasing or decreasing by finding where this derivative is positive or negative.

• If \(x < 30\), then \(30 - x > 0\), indicating that the profit function is increasing.
• If \(x > 30\), then \(30 - x < 0\), showing that the function is decreasing.

Given the specific intervals in the problem, the interval \([0, 20]\) falls entirely within the increasing region, while \([40, 60]\) is within the decreasing region. This confirms that the profit increases on \([0, 20]\) and decreases on \([40, 60]\). This analysis is crucial for businesses to understand how changes in advertising spending impact profits.

Profit Function

A profit function represents the financial gain a company expects from certain business activities, like spending on advertising. It involves mathematical expressions that predict the outcome based on different inputs. Here, the profit function is given as \(P(x) = 200 + 30x - 0.5x^2\), with \(x\) representing the advertising expenses in hundreds of dollars.
Analyzing this function allows businesses to optimize their advertising budget for maximum profit. With the derivative \(P'(x) = 30 - x\), we are equipped to identify critical points where profit is at maximum or needs careful consideration.

• The derivative helps in finding intervals where spending more on advertising will continue to increase profit.
• When the derivative is zero, we might encounter points of maximum profit, crucial for strategic decision-making.

In simple terms, by understanding this function and its derivative, businesses can better navigate their spending strategies to enhance their profitability. By focusing efforts on the increasing intervals, companies can make informed decisions to boost their financial performance effectively.