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Use Lagrange multiplier techniques to find the local extreme values of the given function subject to the stated constraint. If appropriate, determine if the extrema are global. (If a local or global extreme value does not exist enter DNE.) f(x, y)

Respuesta :

Answer:

Explanation:

Given f(x, y) = 5x + y + 2 and g(x, y) = xy = 1

The step by step calculation and appropriate substitution is clearly shown in the attached file.

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The local extreme values for the given function are;

minimum value is 2 - (2√5) while the maximum value is 2 + (2√5)

What is the Lagrange multiplier technique?

We are given the functions;

f(x, y) = 5x + y + 2 and g(x, y) = xy = 1

The general formula for lagrange multiplier is;

L(x, Ξ») = f(x) - Ξ»g(x)

From lagrange multipliers, we know that;

βˆ‡f = Ξ»βˆ‡g Β ----(1)

Since g(x, y) = xy = 1, then;

f_x = Ξ»g_x Β  -----(2)

f_y = Ξ»g_y Β  -----(3)

From eq(2), we have;

Ξ» = 5/y Β  Β ------(4)

From eq 3, we have;

Ξ» = 1/x Β  Β -----(5)

Combining eq 4 and 5 gives us;

5x = y

Put 5x for y into xy = 1 to get;

5xΒ² = 1 and so;

x = ±1/√5

Put ±1/√5 for x in xy = 1 to get;

y = ±√5

Thus, f has extreme values at;

(1/√5, √5), (-1/√5, -√5), (1/√5, -√5), (-1/√5, √5)

At (1/√5, √5), f(x, y) becomes 2 + (2√5)

At (-1/√5, -√5), f(x, y) becomes 2 - (2√5)

At (1/√5, -√5), f(x, y) becomes 2

At (-1/√5, √5), f(x, y) becomes 2

Thus, in conclusion we can say that;

The minimum value is 2 - (2√5) at the point (-1/√5, -√5) while the maximum value is 2 + (2√5) at (1/√5, √5)

Read more about Lagrange Multipliers at; https://brainly.com/question/4609414

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