Ch. 4 - Applications of the Derivative

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 4 - Applications of the Derivative

Problem 4.5.15

Chapter 4, Problem 4.5.15

Minimum sum Find positive numbers x and y satisfying the equation xy = 12 such that the sum 2x + y is as small as possible.

Verified step by step guidance

First, express y in terms of x using the given equation xy = 12. This gives y = 12/x.

Substitute y = 12/x into the expression for the sum, 2x + y, to get a function in terms of x: f(x) = 2x + 12/x.

To find the minimum value of f(x), take the derivative of f(x) with respect to x. Use the derivative rules: f'(x) = d/dx(2x) + d/dx(12/x).

Set the derivative f'(x) equal to zero to find the critical points: f'(x) = 2 - 12/x^2 = 0. Solve for x to find the critical point.

Verify that the critical point found is a minimum by using the second derivative test. Calculate f''(x) and check its sign at the critical point. If f''(x) > 0, the critical point is a minimum.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Optimization

Optimization in calculus involves finding the maximum or minimum values of a function. In this problem, we aim to minimize the sum 2x + y while adhering to the constraint xy = 12. This typically requires the use of techniques such as substitution or the method of Lagrange multipliers to find the optimal values of the variables.

Constraints

Constraints are conditions that must be satisfied in an optimization problem. Here, the equation xy = 12 serves as a constraint that limits the values of x and y. Understanding how to manipulate and incorporate constraints is crucial for finding feasible solutions that meet the problem's requirements.

Derivatives

Derivatives are fundamental in calculus for determining the rate of change of a function. In this context, we can use derivatives to find critical points of the function 2x + y, which will help identify minimum values. By setting the derivative equal to zero, we can solve for x and y that minimize the sum while satisfying the constraint.

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