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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
4장, 문제 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.

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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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:
10:13
Intro to Applied Optimization: Maximizing Area

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.
추천 영상:
10:13
Intro to Applied Optimization: Maximizing Area

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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