Ch. 4 - Applications of Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problema 67

Capitolo 4, Problema 67

Your company can manufacture x hundred grade A tires and y hundred grade B tires a day, where 0 ≤ x ≤ 4 and y = (40 - 10x)/(5-x). Your profit on a grade A tire is twice your profit on a grade B tire. What is the most profitable number of each kind to make?

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First, understand the constraints: you can manufacture between 0 and 400 grade A tires (x) and the number of grade B tires (y) is given by the equation y = (40 - 10x)/(5-x).

Next, note that the profit on a grade A tire is twice that of a grade B tire. Let's denote the profit on a grade B tire as P, then the profit on a grade A tire is 2P.

The total profit function can be expressed as: Total Profit = 2P * x + P * y. Substitute y from the given equation into this profit function.

Simplify the profit function to express it solely in terms of x. This involves substituting y = (40 - 10x)/(5-x) into the profit equation and simplifying.

Finally, determine the value of x that maximizes the profit function. This can be done by taking the derivative of the profit function with respect to x, setting it to zero, and solving for x. Check the endpoints of the interval 0 ≤ x ≤ 4 to ensure you have found the maximum profit.

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Constraints and Feasible Region

In optimization problems, constraints define the limits within which solutions must lie. Here, the constraints are given by the production limits of grade A and grade B tires, specifically 0 ≤ x ≤ 4 and the relationship between x and y. Understanding these constraints helps identify the feasible region where potential solutions exist.

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Profit Function

The profit function represents the total profit earned from producing a certain number of products. In this case, the profit from grade A tires is twice that of grade B tires, which can be expressed mathematically. Formulating the profit function is essential for determining the optimal production levels that maximize profit.

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Optimization Techniques

Optimization techniques, such as finding maximum or minimum values of functions, are crucial in calculus. In this scenario, methods like substitution or the use of derivatives can help identify the production levels of tires that yield the highest profit. Understanding these techniques allows for effective analysis of the profit function within the defined constraints.

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