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Copyright c Oklahoma State University. All rights reserved. Full file at
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258 Solution Guide for Chapter 2 (c) The cannonball reaches its maximum height where the graph reaches a peak. In the right-hand picture below, we see that this occurs when x = 1 . 61 and y = 1 . 39 . Thus the cannonball reaches a maximum height of 1.39 miles at 1.61 miles downrange. 2. Profit : (a) The exercise suggests a horizontal span of 0 to 7 . The table of values on the left below led us to choose a vertical span of - 3 to 6 . The graph is on the right below. (b) From the table of values we see that P (0) = - 2 . This means that, if no widgets are sold, the weekly profit will be - 2 thousand dollars, so the producer will have a loss of $2000 . (c) To find where the largest profit occurs, we locate the peak of the graph. In the figure below, we see that the peak occurs when n = 4 . 83 and P = 5 . 01 . Thus the profit is as large as possible at a sales level of 4 . 83 thousand widgets. Copyright c Oklahoma State University. All rights reserved. Full file at
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SECTION 2.5 Optimization 259 3. Marine fishery : (a) The exercise suggests a horizontal span of 0 to 1 . 5 . The table of values on the left below led us to choose a vertical span of - 0 . 1 to 0 . 1 . The graph is on the right below. (b) The growth rate if the population size is 0 . 24 million tons is represented by G (0 . 24) in functional notation. From a table of values or the graph we see that the value is 0 . 04 million tons per year. (c) From a table of values or the graph we see that G (1 . 42) = - 0 . 02 . This means that if the population size is 1 . 42 million tons then the growth rate is - 0 . 02 million tons per year, so the population is decreasing at a rate of 0 . 02 million tons per year. (d) To find where the largest growth rate occurs, we locate the peak of the graph. In the figure below, we see that the peak occurs when n = 0 . 67 and G = 0 . 07 . Thus the growth rate is the largest at a population size of 0 . 67 million tons. Copyright c Oklahoma State University. All rights reserved. Full file at
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260 Solution Guide for Chapter 2 4. Enclosing a field : (a) The picture should be a rectangle with width marked 3 miles and length marked 5 miles. Since the perimeter of the rectangle is two lengths plus two widths, the enclosure uses 2 × 3 + 2 × 5 = 16 miles of fence, as required. The area of the rectangle is 5 × 3 = 15 square miles. (b) The picture should be a rectangle with width marked 2 miles and length marked 6 miles. Since the perimeter of the rectangle is two lengths plus two widths, the enclosure uses 2 × 2 + 2 × 6 = 16 miles of fence, as required. The area of the rectangle is 6 × 2 = 12 square miles. (c) i. Since there are 16 miles of fence, the width w is somewhere between 0 and 8 miles, and so we use that for a horizontal span. The table below suggests a vertical span of 0 to 20 square miles. In the graph below, width is on the horizontal axis, and area is on the vertical axis. The graph shows the area enclosed by a rectangle of perimeter 16 as a function of the width. From the
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