Maxima and Minima with Lagrange Multipliers and Double Integrals (Math241 Quiz #6) - Prof., Quizzes of Advanced Calculus

The answer key for quiz #6 in math241, covering topics on using lagrange multipliers to find maximum and minimum values of a function subject to a constraint, and calculating double integrals. The problem involves finding the maximum and minimum values of the function x^2y with the constraint x^2 + 2y^2 = 24, and calculating the double integral of the function √(1-r^2) over the region r: 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.

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2010/2011

Uploaded on 07/08/2011

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Math241 D1 Quiz#6 July 7, 2011
Name: Answer Key
1.[10pts] Use Lagrange multipliers to find the maximum and minimum values of the
function subject to the given constraint.
f(x, y) = x2y;x2+ 2y2= 24
Sol) Let g(x, y) = x2+ 2y2.
f=λg, (1)
<2xy, x2>=λ < 2x, 4y > (2)
2xy = 2λx (3)
x2= 4λy (4)
xy =λx
x2= 4λy
x2+ 2y2= 24
The first equation implies that x= 0 or y=λ. If x= 0 then by plugging in the third
equation y=±26. If x6= 0 the y=λ....
2.[10pts] Calculate the double integral
ZZR
1r2rdA, R : 0 r1,0θ2π.
Sol)
ZZR
1r2rdA =Z2π
0Z1
0
1r2rdrdθ (5)
=Z2π
0·1
3(1 r2)3
2¸1
0
(6)
=... (7)
1

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Math241 D1 Quiz#6 July 7, 2011

Name: Answer Key

1.[10pts] Use Lagrange multipliers to find the maximum and minimum values of the

function subject to the given constraint.

f (x, y) = x^2 y; x^2 + 2y^2 = 24

Sol) Let g(x, y) = x^2 + 2y^2.

∇f = λ∇g, (1)

< 2 xy, x 2

= λ < 2 x, 4 y > (2)

2 xy = 2λx (3)

x 2 = 4λy (4)

xy = λx x^2 = 4λy x^2 + 2y^2 = 24

The first equation implies that x = 0 or y = λ. If x = 0 then by plugging in the third

equation y = ± 2

  1. If x 6 = 0 the y = λ....

2.[10pts] Calculate the double integral

∫ ∫

R

1 − r^2 rdA, R : 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π.

Sol)

∫ ∫

R

1 − r^2 rdA =

∫ (^2) π

0

0

1 − r^2 rdrdθ (5)

∫ (^2) π

0

[

(1 − r 2 )

3 2

] 1

0

dθ (6)

1