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