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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 x-coordinate 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