Calculus 2-A11 - Calculus II Calculus: Early Transcendental...

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Calculus II Calculus: Early Transcendental Functions , 4th ed., by Ron Larson, Robert Hostetler, and Bruce H. Edwards (Boston: Houghton Mifflin, 2007; ISBN-10: 0-618-60624-6). Written Assignment 11 Complete the following textbook exercises for sections 13.8, 13.9, and 13.10, and submit them to your mentor for correction and grading. Show all calculations. SECTION 13.8 Exercises Do exercises 2, 6, 10, 46, 54, and 58 on pages 958–959 of the textbook . In Exercises 1– 6, identify any extrema of the function by recognizing its given form or its form after completing the square. Verify your results by using the partial derivatives to locate any critical points and test for relative extrema. Use a computer algebra system to graph the function and label any extrema. g(x,y) = 9 – (x – 3) 2 – (y + 2) 2 ≤ 9 v x,y Relative maximum at (3, -2, 9) gx(x,y) = -2(x – 3), gv(x,y) = 2(y + 2) Critical points gx(x,y) = 0, gv(x,y) = 0 -2(x – 3) = 0 => x=3 -2(y + 2) = 0 => y=-2 For x = 3, y = -2, z = g(x,y) = 9 gxx = -2, gvy = -2, gxy = 0 d = gxx gvy – (gxy) 2 d = (-2)(-2) = 4 > 0 ƒxx = -2 < 0 Relative maximum at (3, -2, 9) = -(x 2 – 4x) – (y 2 – 8y) – 11 = -(x 2 – 4x + 4) – (y 2 – 8y + 16) – 11 + 4 + 16 = -(x – 2) 2 – (y – 4) 2 + 9 ≤ 9 Relative maximum at (2, 4, 9) ƒx = -2x + 4, ƒy = -2y + 8 Critical points ƒx = 0, ƒy = 0 -2x + 4 = 0, -2y + 8 = 0 x = 2, y = 4, z = ƒ(x,y) = 9 Critical point (2, 4) ƒxx = -2, ƒyy = -2, ƒxy = 0
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d = ƒxx ƒyy – (ƒxy) 2 d = (-2)(-2) – 0 = 4 > 0 ƒxx = -2 < 0 Relative maximum at (2, 4, 9) In Exercises 7– 16, examine the function for relative extrema. ƒx = 2x + 6y ƒx = 6x + 20y – 4 Critical points ƒx = 0, ƒy = 0 2x + 6y = 0, 6x + 20y - 4 = 0 x = -6, y = 2, z = ƒ(x,y) = 0 Critical point (-6, 2) ƒxx = 2 > 0, d = ƒxx ƒyy – (ƒxy) 2 d = 40 – 36 = 4 > 0 ƒxx = 4 > 0 Relative maximum at (-6, 2, 0) In Exercises 45– 50, find the critical points and test for relative extrema. List the critical points for which the Second Partials Test fails. ƒx(x, y) = 3x 2 – 12x + 12 => =3(x 2 – 4x + 4) ƒy(x, y) = 3y 2 – 18y + 27 => =3(y 2 – 6y + 9) Critical points ƒx(x, y) = 0, ƒy(x, y) = 0 x 2 – 4x + 4 = 0, y 2 – 6y + 9 = 0 (x – 2) 2 = 0, (y + 3) 2 = 0 x = 2, y = -3 Critical points (2, -3) ƒxx = 6x – 12, ƒyy = 6y + 18, ƒxy = 0 d = ƒxx ƒyy – (ƒxy) 2 =(12 – 12)(-18 + 18) – 0 = 0 d = 0 fails second partial test ƒ(x, y ) = x 3 + y 3 – 6y 2 + 9y 2 + 12x + 27y + 19 = (x 3 – 8 – 6x 2 = 12x) = (y 3 + 27 + 9y 2 + 27y) =(x – 2) 3 + (y + 3) 3
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ƒ(x, y ) ≥ 0 or ≤ 0 dependent on the signs (x – 2) (y + 3) therefore (2, -3, 0) is the saddle point At (2, -3), ƒxxƒyy – (ƒxy) 2 = 0 test fails, (1,-2,0) is a saddle point In Exercises 53– 62, find the absolute extrema of the function over the region ( In each case, contains the boundaries.) Use a computer algebra system to confirm your results. ƒx = 4(2x – y), ƒy = -2(2x – y)
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Calculus 2-A11 - Calculus II Calculus: Early Transcendental...

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