Find the extreme values of the function \(f(x, y) = x^2 + 2y^2\) on the circle \(x^2 + y^2 = 1\).

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Solution:

f(x, y) = x2 + 2y2

such that g(x, y) = x2 + y2 – 1 = 0

Then

\(∇f = 2x \vec{i} + 4y \vec{j}\)

\(∇g = 2x \vec{i} + 2y \vec{j}\)

Then by Lagrange’s multiplier method, for some scalar λ,

∇f = λ∇g

\(2x \vec{i} + 4y \vec{j} = λ (2x \vec{i} + 2y \vec{j}) \)

This gives

2x = λ 2x

λ = 1

and

4y = λ 2y  ⇨ y = 0 [Being λ = 1]

Since we have

g(x, y) = x2 + y2 – 1 = 0

⇨ x2 + 0 – 1 = 0

⇨ x = ± 1

Thus, f attains its extreme at point (1, 0), (-1, 0)

Now,

f(1, 0) = 1 + 0 = 1

f(-1, 0) = 1 + 0 = 1

Thus the extreme value of f is 1.

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