# 3_6 - Chapter 3 Functions and Their Graphs 3.6 1 2...

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Chapter 3 352 Functions and Their Graphs 3.6 Mathematical Models; Constructing Functions 1. V = r 2 h , h = 2 r V r ( 29 = r 2 2 r ( 29 = 2 r 3 2. V = 1 3 r 2 h , h = 2 r V r ( 29 = 1 3 r 2 2 r ( 29 = 2 3 r 3 3. (a) R ( x ) = x - 1 6 x + 100  = - 1 6 x 2 + 100 x (b) R (200) = - 1 6 (200) 2 + 100(200) = - 20,000 3 + 20,000 = 40,000 3 \$13,333 (c) 600 16000 0 0 (d) x = 300 maximizes revenue R (300) = - 1 6 (300) 2 + 100(300) = - 15,000 + 30,000 = \$15,000 = maximum revenue (e) p = - 1 6 (300) + 100 = - 50 + 100 = \$50 maximizes revenue 4. (a) R ( x ) = x - 1 3 x + 100  = - 1 3 x 2 + 100 x (b) R (100) = - 1 3 (100) 2 + 100(100) = - 10,000 3 + 10,000 = 20,000 3 \$6666.67 (c) 300 0 8000 0 (d) x = 150 maximizes revenue R (150) = - 1 3 (150) 2 + 100(150) = - 7500 + 15,000 = \$7500 = maximum revenue (e) p = - 1 3 (150) + 100 = - 50 + 100 = \$50 maximizes revenue

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Section 3.6 Mathematical Models; Constructing Functions 353 5. (a) If x = - 5 p + 100, then p = 100 - x 5 . R ( x ) = x 100 - x 5  = - 1 5 x 2 + 20 x (b) R (15) = - 1 5 (15) 2 + 20(15) = - 45 + 300 = \$255 (c) 100 600 0 0 (d) x = 50 maximizes revenue R (50) = - 1 5 (50) 2 + 20(50) = - 500 + 1000 = \$500 = maximum revenue (e) p = 100 - 50 5 = 50 5 = \$10 maximizes revenue 6. (a) If x = - 20 p + 500, then p = 500 - x 20 . R ( x ) = x 500 - x 20  = - 1 20 x 2 + 25 x (b) R (20) = - 1 20 (20) 2 + 25(20) = - 20 + 500 = \$480 (c) 500 0 4000 0 (d) x = 250 maximizes revenue R (250) = - 1 20 (250) 2 + 25(250) = - 3125 + 6250 = \$3125 = maximum revenue (e) p = 500 - 250 20 = 250 20 = \$12.50 maximizes revenue 7. (a) Let x = width and y = length of the rectangular area. P = 2 x + 2 y = 400 y = 400 - 2 x 2 = 200 - x Then A ( x ) = (200 - x ) x = 200 x - x 2 = - x 2 + 200 x (b) We need x 0 and y 0 200 - x 0 200 x So the domain of A is x 0 x 200 { } (c) 200 10000 0 0 x = 100 yards maximizes area
Chapter 3 Functions and Their Graphs 354 8. (a) Let x = length and y = width of the rectangular field. P = x + 2 y = 3000 y = 3000 - x 2 = 1500 - 1 2 x Then A x ( 29 = 1500 - 1 2 x x = 1500 x - 1 2 x 2 = - 1 2 x 2 + 1500 x (b) 3000 0 1,250,000 0 x = 750 feet maximizes area 9. (a) The distance d from P to the origin is d = x 2 + y 2 . Since P is a point on the graph of y = x 2 - 8, we have: d ( x ) = x 2 + ( x 2 - 8) 2 = x 4 - 15 x 2 + 64 (b) d (0) = 0 4 - 15(0) 2 + 64 = 64 = 8 (c) d (1) = (1) 4 - 15(1) 2 + 64 = 1 - 15 + 64 = 50 = 5 2 7.07 (d) (e) d is smallest when x ≈ - 2.74 and when x 2.74. 10. (a) The distance d from P to (0, –1) is d = x 2 + ( y + 1) 2 . Since P is a point on the graph of y = x 2 - 8, we have: d ( x ) = x 2 + ( x 2 - 8 + 1) 2 = x 2 + x 2 - 7 ( 29 2 = x 4 - 13 x 2 + 49 (b) d (0) = 0 4 - 13(0) 2 + 49 = 49 = 7 (c) d ( - 1) = ( - 1) 4 - 13( - 1) 2 + 49 = 1 - 13 + 49 = 37 6.08 (d) 4 –4 10 0 (e) d is smallest when x ≈ - 2.55 and when x

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