Find the maximum and minimum values of (x ^2 + y^ 2) subject to the condition 5x^2 + 6xy + 5y^2 = 8.
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Find the maximum and minimum values of (x ^2 + y^ 2) subject to the condition 5x^2 + 6xy + 5y^2 = 8.

[From: ] [author: ] [Date: 12-05-19] [Hit: ]
−√1/2), (√2, −√2), (−√2,At (0,At (√2,......
∇f = λ∇g

∇(x² + y²) = λ∇(5x^2 + 6xy + 5y^2)
<2x, 2y> = λ<10x+6y, 6x+10y>
= λ<5x+3y, 3x+5y>

x = λ(5x+3y)
y = λ(3x+5y)
5x² + 6xy + 5y² = 8

x = 5λx+3λy
y = 3λx+5λy

xy = 5λxy + 3λy²
xy = 3λx² + 5λxy

5λxy + 3λy² = 3λx² + 5λxy
λy² = λx²
λ(y² − x²) = 0
λ(x + y)(x − y) = 0

Then either
λ = 0
Or
x = y
Or
x = −y

If λ = 0 then x=y=0

If x = y then x²=y² and
5x² + 6xy + 5y² = 8
5x² + 6x² + 5x² = 8
16x² = 8
x² = 1/2
x = ±√1/2
And y = ±√1/2
(√1/2, √1/2) or (−√1/2, −√1/2)

If x = −y then x² = y² and
5x² + 6xy + 5y² = 8
5x² − 6x² + 5x² = 8
4x² = 8
x² = 2
x = ±√2
(√2, −√2), (−√2, √2)

Test each of these answers
(0,0), (√1/2, √1/2), (−√1/2, −√1/2), (√2, −√2), (−√2, √2)

At (0,0) we have a minimum of z=0
At (√2, −√2) and (−√2, √2) we have a maximum of z=4

At each of the other two, we have an intermediate value, as these were superfluous solutions.
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