Problem 54 - College Algebra and Calculus: An Applied Approach

Question

Optimal Profit The profit \(P\) (in dollars) for a manufacturer of sound systems is given by \(P(x)=-0.0003 x^{2}+150 x-375,000\) where \(x\) is the number of units produced. What production level will yield a maximum profit?

Step-by-Step Solution

Answer

The maximum profit will be attained at a production level of 250,000 units.

Step 1: Establish the Function

The profit function given is \(P(x)=-0.0003x^2 + 150x - 375000\)

Step 2: Compute the Derivative

The derivative of this function is \(P'(x)=-0.0006x + 150\)

Step 3: Find the Critical Point

Setting the derivative equal to zero will provide the peak of the profit function. Compute this as follows: -0.0006x + 150 = 0 leading to x = 250,000 units.

Step 4: Check 2nd Derivative for Conformity

Calculate the second derivative \(P''(x) = -0.0006\) which is always negative. This confirms that x = 250,000 indeed represents the maximum of the original function.

Step 5: Conclusion/Product Recommendation

Therefore, the production level that will yield the maximum profit is 250,000 units.

Key Concepts

Profit Function

A profit function is a mathematical expression that represents the profit gained by producing and selling goods. It is usually denoted by a function of the number of units produced, typically using the variable \(x\). In the problem we have, the profit function is given as \(P(x) = -0.0003x^2 + 150x - 375,000\). This quadratic equation is a typical representation of profit in business calculus. The shape of a quadratic equation is a parabola, and since the coefficient of the \(x^2\) term \(-0.0003\) is negative, it means the parabola opens downwards. The coefficients in the equation have specific meanings:

• The \(x^2\) term: represents the marginal effect of production on profit. The negative sign indicates decreasing returns as production increases, which is common due to factors like material waste or overproduction.


• The linear term (\(150x\)): represents the contribution per unit to the profit. Essentially, this is the profit added by producing one more unit.


• The constant term (-375,000): denotes fixed costs not dependent on production levels, such as rent or salaries.

Understanding the profit function helps to determine at what production level profit is maximized, setting the stage for further calculations.

Derivative Calculation

Calculating derivatives in algebra is like finding the slope of a function at any given point. The derivative of a function gives you the rate at which the function is changing. For profit optimization, it helps us find the points where the rate is zero, which can correspond to maximum, minimum, or points of inflexion. In our exercise, the derivative of the profit function \(P(x)\) is given by:

• \(P'(x) = -0.0006x + 150\)

This derivative expression is obtained by applying the power rule for differentiation. Effectively, we are finding the rate at which profit changes with respect to changes in production levels. The presence of a zero slope in \(P'(x)\) is crucial because it is linked to possible maximum or minimum points on the profit curve, guiding us to make informed decisions about production.

Critical Points

Critical points are the values of \(x\) where the first derivative of the profit function is zero or undefined. These points are significant as they indicate where a function's behavior changes, and in this context, where maximum or minimum profits could occur. For the provided exercise, to find the critical point, we set the derivative equal to zero:

• \( -0.0006x + 150 = 0 \)

Solving for \(x\), we find \(x = 250,000\) units. This is the point where the slope of the function is zero, suggesting a potential peak in the profit curve given our parabolic profit function. By identifying this critical point, we are closer to understanding the optimal production level for maximum profit.

Second Derivative Test

The second derivative test helps determine the nature of the critical points identified from the first derivative. If the second derivative at a particular point is negative, it indicates that the function is concave down at this point, confirming it as a local maximum. If the second derivative is positive, it indicates concavity upwards, pointing to a local minimum.In this problem, the second derivative of the function is:

• \(P''(x) = -0.0006\)

Since \(P''(x)\) is negative for all \(x\), it confirms that the profit function is concave down everywhere. Therefore, the critical point at \(x = 250,000\) units, where we found \(P'(x) = 0\), is indeed a point of maximum profit. This insight into the behavior of the profit function through the lens of calculus is vital for making optimal business decisions.