Question

Find the profit function for the given marginal profit and initial condition. $$ \frac{d P}{d x}=-18 x+1650 \quad P(15)=\$ 22,725 $$

Short answer

The profit function is \(P(x) = -9x^2 + 1650x + C\), where \(C\) is the constant of integration found from the initial condition.

Step-by-step solution

Identify the Marginal Profit Function and Initial Condition

The marginal profit function is \(-18x + 1650\) and the initial condition given is \(P(15) = \$22,725\).

Integrate the Marginal Profit Function

Integrate the function \(-18x + 1650\) with respect to \(x\) to find the profit function. The integral is \(-9x^2 + 1650x + C\), where \(C\) is the constant of integration.

Calculate the Integration Constant using the Initial Condition

Substitute \(x = 15\) and \(P(15) = \$22725\) into the integrated function got from step 2, resulting in the equation \$22725 = -9(15)^2 + 1650(15) + C\). Solve for \(C\) to get the constant of integration.

Write Down the Profit Function

Substitute the constant \(C\) found in step 3 into the integrated function from step 2. This gives the profit function \(P(x) = -9x^2 + 1650x + C\).

Key concepts

Marginal Profit

When a business is analyzing its financial performance, understanding the concept of marginal profit is essential. Marginal profit refers to the additional profit generated by selling one more unit of a product or service. Mathematically, it's the derivative of the profit function, which gives us a rate of change of profit with respect to the number of units sold, denoted by the variable 'x'.

In an algebraic expression, if the profit function is represented by P(x), then the marginal profit would be written as \( \frac{dP}{dx} \). In other words, it tells us how much extra profit is made for every additional unit sold after reaching a certain level. This information is valuable for a business when making decisions on pricing and production levels.

For instance, if a company knows that by increasing production by one unit, they earn an additional \(50 in profit, they can weigh this against the costs needed to produce that extra unit. If the extra cost is less than \)50, it seems like a profitable move to increase production.

Integration of Functions

Integration is a fundamental concept in calculus that represents the summation of areas under a curve, which can also be viewed as the reverse process of differentiation. When we integrate a function, we're basically finding the original function that was differentiated to give us a rate of change or derivative. This original function is typically called the antiderivative.

For a given function f(x), the integration process finds a new function F(x) such that \( \frac{dF}{dx} = f(x) \). The function F(x) can be seen as an accumulation of the quantities represented by f(x) over an interval. In business applications, integrating the marginal profit function, which is the derivative of the profit function, gives us the profit function itself, up to an arbitrary constant known as the constant of integration.

Integration is a powerful tool because it allows us to reconstruct the overall picture from a rate of change. It's like knowing the speed at every point in a journey and then calculating the distance traveled.

Initial Condition in Calculus

In the world of calculus, an initial condition is a piece of additional information that allows us to find a specific solution to a differential equation. In practical terms, it's like having a starting point for a function or a known checkpoint. When we are given a derivative function and we integrate it, the integration process will include a constant of integration because integrals can generate an infinite number of antiderivatives, each differing by a constant.

The use of an initial condition allows us to solve for the exact value of this constant and thus determine the specific function we are interested in. The initial condition will typically be in the form of a point on the original function, such as P(x) at a known x-value.

In problems related to economics or business, an initial condition could represent the known profit at a particular level of production or sales, such as the profit after selling 15 units. By using this known data point, we can accurately determine our constant of integration and, thus, our specific profit function.

Solving for Constants of Integration

Once we integrate a function and obtain an antiderivative with a constant of integration, denoted as 'C', we must find a specific value for this constant to complete the model for a particular situation. The task of finding 'C' is where initial conditions come into play.

To solve for the constant of integration, we use the known values provided by the initial condition. We substitute these values into the integrated function and solve for 'C'. This process is a crucial part of applying calculus to real-world scenarios because it ties the abstract mathematical world of functions and derivatives to actual, measurable quantities.

For example, in the context of the profit function exercise, after integrating the marginal profit function, we were left with a constant 'C'. To find its value, we used the initial condition that the profit was $22,725 after selling 15 units. Placing these numbers into our integrated function allowed us to solve for 'C', giving us the precise profit function that could then be applied to predict profits at different levels of production or sales.

Having such a function is invaluable for a business as it can use this to forecast future profits, set targets, and make strategic decisions about resource allocation and product pricing strategies.