9. p(x) = x 3 8x 2 5x p(x) = x 3 + 3x 2 33x p(x) = x x p(x) = x 3 + 5x x p(x) = x 4 50x

Transcription

1 Section 6.3 Etrema and Models Eercises In Eercises 1-8, perform each of the following tasks for the given polnomial. i. Without the aid of a calculator, use an algebraic technique to identif the zeros of the given polnomial. Factor if necessar. ii. On graph paper, set up a coordinate sstem. Label each ais, but scale onl the -ais. Use the zeros and the end-behavior to draw a rough graph of the given polnomial without the aid of a calculator. iii. Classif each local etrema as a relative minimum or relative maimum. Note: It is not necessar to find the coordinates of the relative etrema. Indeed, this would be difficult without a calculator. All that is required is that ou label each etrema as a relative maimum or minimum. 1. p() = ( + 6)( 1)( 5) 2. p() = ( + 2)( 4)( 7) 3. p() = p() = p() = p() = p() = p() = In Eercises 9-16, perform each of the following tasks for the given polnomial. i. Use a graphing calculator to draw the graph of the polnomial. Adjust the viewing window so that the etrema or turning points of the polnomial are visible in the viewing window. Cop the resulting image onto our homework paper. Label and scale each ais with min, ma, min, and ma. ii. Use the maimum and/or minimum utilit in our calculator s CALC menu to find the coordinates of the etrema. Label each etremum on our homework cop with its coordinates and state whether the etremum is a relative or absolute maimum or minimum. 9. p() = p() = p() = p() = p() = p() = p() = p() = A square piece of cardboard measures 12 inches per side. Cherie cuts four smaller squares from each corner of the cardboard square, tossing the material aside. She then bends up the sides of the remaining cardboard to form an open bo with no top. Find the dimensions of the squares cut from each corner of the original piece of cardboard so that 1 Coprighted material. See:

2 594 Chapter 6 Polnomial Functions Cherie maimizes the volume of the resulting bo. Perform each of the following steps in our analsis. a) Set up an equation that determines the volume of the bo as a function of, the length of the edge of each square cut from the four corners of the cardboard. Include an pictures used to determine this volume function. b) State the empirical domain of the function created in part (a). Use our calculator to sketch the graph of the function over this empirical domain. Adjust the viewing window so that all etrema are visible in the viewing window. c) Cop the image in our viewing window onto our homework paper. Label and scale each ais with min, ma, min, and ma. Use the maimum utilit to find the coordinates of the absolute maimum on the function s empirical domain. d) What are the measures of the four squares cut from each corner of the original cardboard? What is the maimum volume of the bo? 18. A rectangular piece of cardboard measures 8 inches b 12 inches. Schuler cuts four smaller squares from each corner of the cardboard square, tossing the material aside. He then bends up the sides of the remaining cardboard to form an open bo with no top. Find the dimensions of the squares cut from each corner of the original piece of cardboard so that Schuler maimizes the volume of the resulting bo. Perform each of the following steps in our analsis. a) Set up an equation that determines the volume of the bo as a function of, the length of the edge of each square cut from the four corners of the cardboard. Include an pictures used to determine this volume function. b) State the empirical domain of the function created in part (a). Use our calculator to sketch the graph of the function over this empirical domain. Adjust the viewing window so that all etrema are visible in the viewing window. c) Cop the image in our viewing window onto our homework paper. Label and scale each ais with min, ma, min, and ma. Use the maimum utilit to find the coordinates of the absolute maimum on the function s empirical domain. d) What are the measures of the four squares cut from each corner of the original cardboard? What is the maimum volume of the bo? 19. Restrict the graph of the parabola = 4 2 /4 to the first quadrant, then inscribe a rectangle inside the parabola, as shown in the figure that follows. (, ) a) Epress the area of the inscribed rectangle as a function of. b) State the empirical domain of the

3 Section 6.3 Etrema and Models 595 function defined in part (a). Use our calculator to graph the area function over its empirical domain. Adjust the window parameters so that all etrema are visible in the viewing window. c) Cop the image in our viewing window to our homework paper. Label and scale each ais with min, ma, min, and ma. Use the maimum utilit to find the coordinates of the absolute maimum on the function s empirical domain. Label our graph with this result. utilit to find the coordinates of the absolute maimum on the function s empirical domain. Label our graph with this result. d) What are the length of the base and height of the triangle of maimum area? d) What are the dimensions of the rectangle of maimum area? 20. Restrict the graph of the parabola = 4 2 /4 to the first quadrant, then inscribe a triangle inside the parabola, as shown in the figure that follows. (, ) a) Epress the area of the inscribed triangle as a function of. b) State the empirical domain of the function defined in part (a). Use our calculator to graph the area function over its empirical domain. Adjust the window parameters so that all etrema are visible in the viewing window. c) Cop the image in our viewing window to our homework paper. Label and scale each ais with min, ma, min, and ma. Use the maimum

4 Chapter 6 Polnomial Functions 6.3 Solutions 1. Set p() = ( + 6)( 1)( 5) equal to zero and use the zero product propert to identif zeros at = 6, 1, and 5. Hence, the graph of p will have -intercepts at ( 6, 0), (1, 0), and (5, 0). In addition, if ou were to epand p() = (+6)( 1)( 5), the leading term would be 3, so the graph of p will have to rise from negative infinit, wiggle through the -intercepts, then rise to positive infinit. Consequentl, the graph will have to look somewhat like what follows. Local Maimum ( 6,0) (1,0) (5,0) Local Minimum The relative etrema are classified on the graph above. 3. First, factor p() = b grouping, then complete the factorization b using the difference of squares pattern. p() = 2 ( 6) 4( 6) p() = ( 2 4)( 6) p() = ( + 2)( 2)( 6) Set p() = ( + 2)( 2)( 6) equal to zero and use the zero product propert to identif zeros at = 2, 2, and 6. Hence, the graph of p will have -intercepts at ( 2, 0), (2, 0), and (6, 0). In addition, the leading term of p() is 3, so the graph of p will have to rise from negative infinit, wiggle through the -intercepts, then rise to positive infinit. Consequentl, the graph will have to look somewhat like what follows. Local Maimum ( 2,0) (2,0) (6,0) Local Minimum The relative etrema are classified on the graph above.

5 Section 6.3 Etrema and Models 5. First, factor out the gcf ( in this case) from p() = , then complete the factorization using the ac-method. p() = [ ] p() = [ ] p() = [(2 7) + 6(2 7)] p() = ( + 6)(2 7) Set p() = ( + 6)(2 7) equal to zero and use the zero product propert to identif zeros at = 0, 6, and 7/2. Hence, the graph of p will have -intercepts at (0, 0), ( 6, 0), and (7/2, 0). In addition, the leading term of p() is 2 3, so the graph of p will have to rise from negative infinit, wiggle through the -intercepts, then rise to positive infinit. Consequentl, the graph will have to look somewhat like what follows. Local Maimum ( 6,0) (0,0) (7/2,0) Local Minimum The relative etrema are classified on the graph above. 7. First, factor out the gcf ( 2 in this case) from p() = , then complete the factorization using the ac-method. p() = 2[ ] p() = 2[ ] p() = 2[( 7) + 5( 7)] p() = 2( + 5)( 7) Set p() = 2(+5)( 7) equal to zero and use the zero product propert to identif zeros at = 0, 5, and 7. Hence, the graph of p will have -intercepts at (0, 0), ( 5, 0), and (7, 0). In addition, the leading term of p() is 2 3, so the graph of p will have to fall from positive infinit, wiggle through the -intercepts, then fall to negative infinit. Consequentl, the graph will have to look somewhat like what follows.

6 Chapter 6 Polnomial Functions Local Maimum ( 5,0) (0,0) Local Minimum (7,0) The relative etrema are classified on the graph above. 9. The maimum and minimum utilities in the CALC menu were used to find the local maimum and minimum values of p() = shown in (c) and (d). (a) (b) (c) (d) There is a local maimum at( , ) and a local minimum at ( , ) Answers ma differ slightl due to roundoff error. 11. The maimum and minimum utilities in the CALC menu were used to find the local maimum and minimum values of p() = shown in (c) and (d). (a) (b) (c) (d) There is a local maimum at ( , ) and a local minimum at ( , ) Answers ma differ slightl due to roundoff error. 13. The maimum and minimum utilities in the CALC menu were used to find the local maimum and minimum values of p() = shown in (c), (d), and (e). (a) (b)

7 Section 6.3 Etrema and Models (c) (d) (e) There are absolute minima at at ( 5, 576) abd (5, 576) and a local maimum at (0, 49). Answers ma differ slightl due to roundoff error. 15. The maimum and minimum utilities in the CALC menu were used to find the local maimum and minimum values of p() = shown in (c), (d), and (e). (a) (b) (c) (d) (e) There is an absolute minimum at ( , ), a local maimum at ( , ), and a local minimum at ( , ). Answers ma differ slightl due to roundoff error. 17. a) Let represent the length (in inches) of each side of four corners cut from the cardboard square, as shown in (a). Because two square are cut from each side, that leaves 12 2 inches on a side. When we cut out the squares and throw them awa, then bend up the sides, we get the bo with dimensions shown in (b).

8 Chapter 6 Polnomial Functions (a) (b) The volume of a bo is given b multipling the length and width of the base times the height of the bo. Hence, the volume V as a function of, is given b or equivalentl, V = (12 2)(12 2), V = (12 2) 2. b) There are 12 inches on a side. We have two cut two squares with sides of length from each side. Thus, the smallest value of is zero and the largest possible value of is 6. Consequentl,the empirical domain is [0, 6]. We use our calculator to plot v = (12 2) 2 over this domain [0, 6] and then use the maimum utilit on the CALC menu to find the maimum value of V on the empirical domain [o, 6]. (c) (d) (e)

9 Section 6.3 Etrema and Models c) We record the results on our homework as follows. V 200 (2,128) V ()=(12 2) d) Cut squares 2 inches on a side to produce a bo having volume 128 in a) Pictured below is the graph of = 4 2 /4 in the first quadrant. A point (, ) lies on the graph and a rectangle is formed. Note that the width of the rectangle is. The height of the rectangle is. (, ) Therefore, the area of the triangle is given as a function of and b A =. However, this is a function of two variables. We need to eliminate one of the variables. This is eas to do because the point (, ) is on the graph of = 4 2 /4. Replace the in A = with = 4 2 /4 to obtain the area as a function of alone. A = (4 2 /4) b) Note that the -intercept of the parabola = 4 2 /4 pictured above is (4, 0). You can also determine the -intercept b setting 0 = 4 2 /4 and solving for (we leave that for ou). Because the point (, ) is constrained to lie on the parabola in

10 Chapter 6 Polnomial Functions the first quadrant (as shown above), this forces to between zero and four. Hence, the empirical (practical) domain is [0, 4]. We use our graphing calculator to draw the graph of A = (4 2 /4) on the empirical domain [0, 4], as shown in (a), (b), and (c). We used the maimum utilit in the CALC menu to determine the maimum shown in (c). Answers ma var due to roundoff error. (a) (b) (c) c) Cop the image onto our homework. A 10 ( , ) V ()=(4 2 /4) Note that we have an absolute maimum at ( , ). d) The area will be a maimum when the width of the rectangle is = To find the height of the rectangle, substitute = into = 4 2 /4 to obtain = 4 ( ) 2 /