These extreme values occur either at endpoints or critical points. Īs mentioned earlier, since A A is a continuous function on a closed, bounded interval, by the extreme value theorem, it has a maximum and a minimum. Maximize A ( x ) = 100 x − 2 x 2 A ( x ) = 100 x − 2 x 2 over the interval. Therefore, we consider the following problem: If the maximum value occurs at an interior point, then we have found the value x x in the open interval ( 0, 50 ) ( 0, 50 ) that maximizes the area of the garden. Therefore, let’s consider the function A ( x ) = 100 x − 2 x 2 A ( x ) = 100 x − 2 x 2 over the closed interval. However, we do know that a continuous function has an absolute maximum (and absolute minimum) over a closed interval. We do not know that a function necessarily has a maximum value over an open interval. Therefore, we are trying to determine the maximum value of A ( x ) A ( x ) for x x over the open interval ( 0, 50 ). Therefore, we need x > 0 x > 0 and y > 0. To construct a rectangular garden, we certainly need the lengths of both sides to be positive. A ( x ) = x īefore trying to maximize the area function A ( x ) = 100 x − 2 x 2, A ( x ) = 100 x − 2 x 2, we need to determine the domain under consideration. Let’s look at how we can maximize the area of a rectangle subject to some constraint on the perimeter. However, what if we have some restriction on how much fencing we can use for the perimeter? In this case, we cannot make the garden as large as we like. Certainly, if we keep making the side lengths of the garden larger, the area will continue to become larger. For example, in Example 4.32, we are interested in maximizing the area of a rectangular garden. However, we also have some auxiliary condition that needs to be satisfied. We have a particular quantity that we are interested in maximizing or minimizing. The basic idea of the optimization problems that follow is the same. Solving Optimization Problems over a Closed, Bounded Interval In this section, we show how to set up these types of minimization and maximization problems and solve them by using the tools developed in this chapter. In manufacturing, it is often desirable to minimize the amount of material used to package a product with a certain volume. For example, companies often want to minimize production costs or maximize revenue. One common application of calculus is calculating the minimum or maximum value of a function. 4.7.1 Set up and solve optimization problems in several applied fields.The worksheets can be used as a test of mastery before moving on to subsequent video lessons in the series.Įvery problem in the worksheets comes with a fully worked step-by-step written solution and answer key. Every worksheet consists of problems that directly follow from what was learned in the video lessons. Watch the video lesson to learn the concept, then work these worksheets to test skills. These worksheets are to be used along with the Calculus 1 Tutor video lessons. Worksheet 17: Improper Integrals (12 Problems) Worksheet 16: Integration by Trig Substitution (10 Problems) Worksheet 15: Integration by Parts (20 Problems) Worksheet 14: Derivatives and Integrals of Logarithms (20 Problems) Worksheet 13: Derivatives and Integrals of Exponentials (21 Problems) Worksheet 12: Calculating Volume with Integrals (16 Problems) Worksheet 11: Integration by Substitution (20 Problems) Worksheet 10: Solving Integrals (20 Problems) Worksheet 9: Introduction to Integrals (20 Problems) Worksheet 8: Curve Sketching Using Derivatives (8 Problems) Worksheet 6: Higher Order Derivatives (21 Problems) Worksheet 5: The Chain Rule (23 Problems) Worksheet 4: Derivatives of Trig Functions (20 Problems) Worksheet 3: Differentiation Formulas (24 Problems) Worksheet 2: The Derivative Defined as a Limit (12 Problems) Worksheet 1: What is a Derivative (14 Problems) These Worksheets are Available for Download
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