Question

Let f(x, y) = x2 6y2. find the maximum and minimum values of f subject to the given constraint?

Answer 1

The minimum and maximum values are (16/9)√3  and -(16/9)√3.

According to the statement

we have given that the function

First of all, since the constraint's graph is a circle, which is a closed loop, and f is continuous in $R^2$, there must exist a constrained global maximum and minimum value.

Lagrange Multipliers:

f(x,y) = xy^2

g(x,y) = x^2 + y^2

We want ∇f = λ∇g so we get the following system of equations

1. y^2 = 2λx

2. 2xy = 2λy

3. x^2 + y^2 = 4 ← The constraint.

Equation 2 implies that y = 0 or λ = x

We can ignore y = 0, since that will make f(x,y) = 0 and clearly f(x,y) takes on both positive and negative values subject to the constraint.

Plugging in the alternative, λ = x to equation 1 gives y^2 = 2x^2.

Plugging this into the constraint gives 3x^2 = 4 so that x^2 = 4/3 and y^2 = 8/3

Taking square roots gives

x = ±√(4/3) = ±(2/3)√3

y = ±√(8/3) = ±(2/3)√6

At the points < (2/3)√3 , ±(2/3)√6 >, f(x,y) = (16/9)√3 ← Maximum

At the points < -(2/3)√3 , ±(2/3)√6 >, f(x,y) = -(16/9)√3 ← Minimum

So, The minimum and maximum values are (16/9)√3  and -(16/9)√3.

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