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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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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
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
0
dθ (6)
1