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Unformatted text preview: Lecture 27 Optimization problems (Relevant section from Stewart, Sixth Edition: 4.7) We now enter – albeit briefly – the territory of “Applied Calculus,” the use of Calculus to solve some “practical” problems. These problems involving the maximizing or minimizing of something, i.e., a quantity “ Q ”, for example, a distance, a time, an area, a volume, or perhaps the cost of constructing something (which, in turn, may depend upon its area or volume). In real life, the quantities Q that one seeks to minimize or maximize will involve many variables. This will be the subject of a future calculus course on functions of several variables (e.g. MATH 237 or MATH 227). Since this course deals with functions of a single real variable, we’ll examine some problems that can be transformed into such functions. Indeed, this is often the stumbling block – how to transform a problem with a number of components into one with a single variable. It is recommended that you read Section 4.7 of Stewart’s textbook thoroughly for a good under standing of this topic. It begins with a helpful outline of the strategy behind solving optimization problem. A number of illustrative problems are then solved. In this lecture, we have time only to consider a few problems, starting with a simple one and then working our way up to a very interesting problem from Physics: the refraction of light. Example 1: Find the area of the largest rectangle that can be inscribed in a semicircle of radius R . The first step is to sketch a picture of the situation, if only to get a feel for the problem. y = √ R 2 x 2 x R R b h y 193 The next step is to introduce a quantity that could serve as the unifying variable in the problem. In this case, we may wish to consider the xcoordinate of the side of the rectangle, as indicated in the figure above. The domain of this x variable will be [0 ,R ]. By symmetry, this implies that the rectangle will extend from x to x . As such, its base length b is b = 2 x. (1) The height h of the rectangle will then be determined by the semicircle: h = radicalbig R 2 x 2 . (2) The area of the rectangle now becomes a function of x : A = bh = 2 x radicalbig R 2 x 2 = A ( x ) . (3) The goal is now to maximize A ( x ) over the interval [0 ,R ]. Note that A (0) = 0 , A ( R ) = 0 , (4) as expected from the sketch: When x = 0, the rectangle becomes a vertical line from (0 , 0) to (0 ,R ). When x = R , the rectangle becomes the horizontal line y = 0 and R ≤ x ≤ R . We look for critical points of A ( x ): A ′ ( x ) = 2 radicalbig R 2 x 2 + 2 x · 1 2 · 2 x √ R 2 x 2 = 2 radicalbig R 2 x 2 2 x 2 √ R 2 x 2 = 2 R 2 x 2 x 2 √ R 2 x 2 = 2 R 2 2 x 2 √ R 2 x 2 . (5) We see that A ′ ( x ) = 0 ⇒ 2 x 2 = R 2 ⇒ x 2 = R 2 2 ⇒ x = R √ 2 . (6) Evaluating A at this critical point, we see that A parenleftbigg R 2 √ 2 parenrightbigg = 2 parenleftbigg R √ 2 parenrightbiggparenleftbigg R √ 2 parenrightbigg = R 2 . (7) Since this value is greater than the value of...
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 Fall '08
 SPEZIALE
 Calculus, Derivative, Continuous function

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