2.3 Maximum and Minimum Applications

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2 Section Maximum and Minimum Applications Maximizing (or minimizing) is an important technique used in various fields of study. In business, it is important to know how to find the maximum profit and how to minimize loss. In physics, it is important to find values such as the maximum speed of an object, or the minimum distance between two moving bodies. In Calculus, maximization is done by analyzing derivatives of functions, a topic beyond our reach now. However, we can look at these problems numerically, using graphing calculators and the properties of the graph of an equation to find the maximum or minimum values. This section will deal with several different types of these problems. We start off with max/min problems of a type of that should be familiar to you. These are the ones involving a quadratic equation: y = ax + bx + c. Example 1 (Quadratic equations) The height above ground of an object at time t (in seconds) traveling vertically and subject only to the pull of gravity is given by the equation h = 16t + vot + h o, where ho is the initial height (in feet) and v o is the initial velocity (in feet per second). A bullet fired upward from ground level has an initial velocity of 480 feet per second. How long does it take to reach its maximum height? How high does it go? Solution: Here are two ways to solve this problem. I. Since the equation h = 16t + 480t ( v o = 480, ho = 0) is quadratic, we know the graph of the equation has the following properties: A. The points on the graph are the ordered pairs (t, h), where h is the height at the time t. B. The graph is a parabola which will have a vertex. C. Since a = 16 is negative, the graph opens downward. D. Since B and C are true, the vertex must be the highest point on the graph (maximum). Both questions will be answered if we find the ordered pair for the vertex. b This can be done by completing the square or by using the fact that t = a at the vertex. From this formula we find t = = = 15. This ( 16) 3 tells us that it takes 15 seconds for the bullet to reach its maximum height. The maximum height is then found by finding h when t is 15. So h = 16(15) + 480(15) = The maximum height is 3600 feet.

3 156 II. We can also solve this by graphing the equation h = 16t + 480t and finding the maximum using a calculator. Here are the steps to accomplish this task. A. Graph Y 16x 1 = + 480x on the graphing calculator. Remember that X is t and Y1 is h. B. We must adjust the viewing window to see all the important features of the graph. C. Setting the window as below we get the following graph. On CD: Graphing Polynomials D. To find the maximum (which is the vertex), we use the maximum finder and get x = 15 and y = On CD: Finding the Max and Min of Polynomials (Note: When you use a calculator to solve a problem you are introducing round-off errors. The resulting answers are usually not exact.) It takes 15 seconds for the bullet to reach a maximum height of 3600 feet. Now we will look at some examples that deal with cubic and other equations. Note that at this time, we can solve these problems only by using the graphing calculator. Algebraic approaches are introduced in Calculus.

4 Section Example (Maximizing volume) An open-top box is to be made from a sheet of cardboard that measures 0 inches by 35 inches by cutting squares of equal size from each of the corners and then folding the flaps up. What size squares should be removed to maximize the volume of the box? Solution: x 35 x 0 x L W This problem involves finding the volume of a box so we need to use the formula for the volume: V = LWH. A. What are the length (L), width (W), and height (H) for our box? Well, the length of the original sheet of cardboard is 35 inches. In order to make the box, we will be removing a square of length x from each corner and folding along the dotted lines. After doing so, we have that L = 35 x x = 35 x. The same process takes place for the width so we have W = 0 x. The height of the box is determined by the dimensions of the squares that were removed so H = x. Then V = ( 35 x)(0 x) x. B. The volume must now be maximized. To do this, we graph the equation in an appropriate window. Note that Y 1 is the volume and x is the side length of the squares. (See comments below.)

5 158 C. Use the maximum finder to find the maximum volume for t he box. When using the maximum finder, we get that the maximum volume is approximately , which is the y-value. The size of the squares that are removed is approximately 4.098, which is the x-value. Since we were asked to find the size of the squares, we have that they are inches by inches. [Co mments: 1. How did we know to let x go from 0 to 10 in the viewing window? In order to construct a box in this fashion we must remove less than 10 inches of material from each corner. If we removed mo re than 10 inches then we wouldn t have a box! Finding usable values for Ymin and Ymax will then require that you look at the table, or trace along the graph.. The equation that we have for the volume is not a quadratic so the algebraic technique used in example 1 does not apply.] Example 3 (Maximizing area) Given the equation y = 8 x, find the positive value of x for which the triangle with vertices at the points (0, 0), (x, 0), and (x, y) has maximum area. Solution: (x, y) (0,0) (x, 0) We need to use the formula for the area of a triangle A = 1 bh.

6 Section A. What are the base b and the height h for our problem? In this case 1 the base is x and the height is y. So the area is A = xy. Since ( x, y) is on the graph of y = 8 x, the formula for the area becomes 1 A = x( 8 x ). B. Now we graph the equation in an appropriate window. We choose the values for x to go from - to 5 since the actual values for x must be between 0 and 8. Note that Y here is the area A. 1 C. Use the calculator to find that the maximum y-value for this graph. Remember, the y-value represents the area. So an x-value of approximately will give you a triangle with maximum area of approximately [Comments: Even though the original equation was a quadratic, the equation that we found for the area was not quadratic. Since 1 x( 8 x ) is not quadratic, the algebraic formula for the vertex does not apply.]

7 160 Example 4 ( Minimizing distance) Find the minim um distance from the graph of y = x 3 3x and the point (1, 1). Solution: We need to find the formula for the distance from the point (1, 1) to all points on the graph. A. The points on the graph are the ordered-pairs (x, y) which satisfy y x 3 3x. Then a typical point on the graph is x, x 3 3x. = ( ) B. To find the distance d between the points (1, 1) and ( x, x 3x) use the distance formula. This says that 3 ( x 1) + ( x 3 1) d = x. This is the quantity we want to 3, we minimize. C. Graph the equation in the appropriate window, where Y 1 is the distance d given by the formula above. Then y here represents the distance from the point (1, 1) to the original graph. D. Use the calculator to find the minimum distance. Notice that this graph has two candidates for the minimum value. We are concerned with finding the absolute minimum distance. After some calculator work, we see that this occurs at the low point on the right side of the picture.

8 Section So we have that when x = (approx.), the minimum distance d is approximately The point on the original graph which is closest to (1, 1) is now determined by plugging the x-value into the 3 original equation y = x 3x. The resulting closest point is approximately (1.864, 0.884). Check the original graph to see if the answer makes sense. Here s a more involved example for you. Multipart problems like these often appear in Calculus courses. Example 5 (Minimizing time) A certain lifeguard can run at 15.7 feet per second and can swim at 5.9 feet per second. He is located 0 feet away from the edge of the ocean. A swimmer is drowning 50 feet out to sea. If the distance along the shore between the lifeguard and the drowning victim is 40 feet, where should the lifeguard enter the water so that he minimizes the time that it takes to reach the drowning swimmer? Solution: S water 50 feet A x E B 0 feet 40 - x L land This problem requires us to combine several of the ideas used earlier. To set up the problem let s examine the picture and label everything. Let A be the point on the shore closet to the lifeguard L, and let B be the point on the sho re closest to the swimmer S. Let E be the point on the shore where the lifeguard will enter the water. We are given that the distance between A and B is 40 feet. Let x be the distance from A to E. The distance from E to B is then 40 x. We need to find a formula in terms of x for the amount of time it will take for the lifeguard to reach the swimmer. We use the standard D fo rmula D = RT, in the form T =, to find expressions for the time he R spends running and for the time that he spends swimming. The total time will be the sum of these two expressions.

9 16 A. Time spent running: His speed (rate) is He runs an unknown distance d from point L to point E. The Pythagorean Theorem tells us that d = 0 + x, so that d = 0 + x. Then the time he d 0 + x spends running from L to E is =. r 15.7 B. Time spent swimming: His speed is 5.9 and the distance from E to S is ( 40 x ) + 50 ( 40 x). Then the time that he spends swimming is C. Combine these to find the total time, expressed in terms of x. ( 40 x) x 50 T = value is the time T at a minimum? ( ) 50. The question is: for which x 0 + x 40 x + D. Graph Y 1 = + on the graphin g calculator and use the minimum finder to obtain x = and y = So he should enter the water approximately 4.7 feet down shore to reach the swimmer in a minimum time of approximately seconds. In all of these problems the general strategy is to first find a formula for the quantity that you want to maxim ize or minimize. This sho uld be written in terms of a single unknown quantity. This step is the most difficult aspect of these problems. If the formula (equation) is quadratic then you can find the vertex algebraically. However, you can always approximate the max/min by using a graphing calculator. It is im portant to know what the two different variables of the ordered pairs on the graph represent, and what you are asked to find. It is easy to get confused in these problems and to misinterpret the values.

10 Section Exercises. 3 Find the vertex of the following quadratic equations and state whether it is a maximum or a minimum. 1. y = 15x 3x + 7. y = 5 x + x y = 0.01x 0.3x y = x x Find the maximum or minimum value for y given the restrictions on x. (Round your answers to decimal places.) 3 5. y = x x 6x a. 0 < x < 3 Minimum b. < x < 0 Maximum 6. y = x 3 + 8x 145x 450 a. 15 < x < 3 Maximum b. 0 < x < 10 Minimum Solve the following problems. For problems 7 and 8, use the height equation from Example A ball is thrown upward from the top of a 96-foot-high tower with an initial velocity of 80 feet per second. When does the ball reach its maximum height and how high is it at that time? 8. A ball is thrown upward from a height of 6 feet with an initial velocity of 3 feet per second. Find its maximum height. 9. During the Civil War, the standard heavy gun for coastal artillery was the 15-inch Rodman cannon, which fired a 330-pound shell. If one of these guns is fired from the top of a 50-foot-high shoreline embankment, then the height of the shell above the water (in feet) can be approximated by the equation h = x +. 3x + 50, where x is the horizontal distance (in feet) from the foot of the embankment to a point directly under the shell. How high does the shell go, and how far away does it hit the water? (Hint: How high will it be when it hits the water?) 10. Based on data from past years, a consultant informs Tim s Bicycles that their profit from selling x bicycles is given by the equation x p = 50x How much profit do they make by selling bicycles? How many bicycles should be sold to maximize the profit? 11. An auto parts manufacturer makes radiators that sell for $350 each. The profit generated by selling x radiators is approximated by the equation P =.01x + 375x 600,000. What number of radiators will produce the largest possible profit? What is the largest possible profit?

11 A farmer has 500 feet of fencing and wants to fence in a rectangular area next to a straight river. What are the dimensions of the field with largest possible area, assuming that no fencing is needed along the river? 13. An open-top rectangular box is to be made from a 35-inch by 35-inch piece of sheet metal by cutting out equal size squares from the corners and folding up the sides. What size squares should be removed in order to produce a box with maximum volume? 14. An open-top box with a square base is to be constructed from 150 sq cm of material. What dimensions for the box will produce the largest possible volume? (Hint: You need to find a formula for the height of the box given that the surface area is 150 sq cm.) 15. Find the minimum distance from the graph of y = x to the point (3, 0). 16. Find the minimum distance from the graph of y = x + x + 6 to the point (1, ). 17. For every point (x, y) which is in the first quadrant and is on the graph y = 4 4x, consider the rectangle with corners ( x, 0), ( x, y), (x, 0), and (x, y). For which value of x does this rectangle have maximal area? helicopter 3 1 mile miles Anne 18. Anne is standing on a straight road and wants to reach her helicopter, which is located miles down road from her, then a mile to the right in a field. She can run 5 miles per hour on the road and 3 miles per hour in the field. She plans to run down the road, then cut diagonally across the field to reach the helicopter. Where should she leave the road in order to reach the helicopter in exactly 4 minutes? What is the least amount of time that it will take her to reach the helicopter? (Hint: See example 4.)