SciVoyage

Location:HOME > Science > content

Science

Optimizing the Function f(x, y) xy on an Elliptical Constraint Using the Method of Lagrange Multipliers

January 07, 2025Science4288
Optimizi

Optimizing the Function f(x, y) xy on an Elliptical Constraint Using the Method of Lagrange Multipliers

In this article, we will explore the process of finding the maximum and minimum values of the function f(x, y) xy subject to the elliptical constraint 3x^2 y^2 3. This problem can be efficiently solved using the method of Lagrange multipliers.

Step-by-Step Solution

Step 1: Setting Up the Problem

We aim to maximize the function f(x, y) xy under the constraint g(x, y) 3x^2 y^2 - 3 0. The method of Lagrange multipliers involves introducing a multiplier λ to balance the function and constraint equations.

Step 2: Computing the Gradients

The gradients of the function f(x, y) xy and the constraint g(x, y) 3x^2 y^2 - 3 are calculated as follows:

nabla f (y, x)

nabla g (6x, 2y)

Step 3: Setting Up the Lagrange Multiplier Equations

According to the method of Lagrange multipliers, we have:

nabla f λ nabla g

This leads to the following system of equations:

y λ 6x

x λ 2y

3x^2 y^2 3

Step 4: Solving the System of Equations

From the first two equations, we can express λ as follows:

λ frac{y}{6x}

λ frac{x}{2y}

Equating these expressions for λ gives:

frac{y}{6x} frac{x}{2y}

Cross-multiplying yields:

2y^2 6x^2 implies y^2 3x^2

Step 5: Substituting into the Constraint

Substituting y^2 3x^2 into the constraint 3x^2 y^2 3 results in:

3x^2 3x^2 3 implies 6x^2 3 implies x^2 frac{1}{2} implies x pm frac{1}{sqrt{2}}

Substituting back to find y gives:

y^2 3x^2 3 cdot frac{1}{2} frac{3}{2} implies y pm frac{sqrt{3}}{sqrt{2}} pm frac{sqrt{6}}{2}

Step 6: Finding Critical Points

The critical points are:

left(frac{1}{sqrt{2}}, frac{sqrt{6}}{2}right)

left(frac{1}{sqrt{2}}, -frac{sqrt{6}}{2}right)

left(-frac{1}{sqrt{2}}, frac{sqrt{6}}{2}right)

left(-frac{1}{sqrt{2}}, -frac{sqrt{6}}{2}right)

Step 7: Evaluating f(x, y) xy at Critical Points

The function values at these points are found as follows:

fleft(frac{1}{sqrt{2}}, frac{sqrt{6}}{2}right) frac{1}{sqrt{2}} cdot frac{sqrt{6}}{2} frac{sqrt{3}}{2}

fleft(frac{1}{sqrt{2}}, -frac{sqrt{6}}{2}right) frac{1}{sqrt{2}} cdot -frac{sqrt{6}}{2} -frac{sqrt{3}}{2}

fleft(-frac{1}{sqrt{2}}, frac{sqrt{6}}{2}right) -frac{1}{sqrt{2}} cdot frac{sqrt{6}}{2} -frac{sqrt{3}}{2}

fleft(-frac{1}{sqrt{2}}, -frac{sqrt{6}}{2}right) -frac{1}{sqrt{2}} cdot -frac{sqrt{6}}{2} frac{sqrt{3}}{2}

Step 8: Conclusion

The maximum value of f(x, y) xy is frac{sqrt{3}}{2} and the minimum value is -frac{sqrt{3}}{2} under the given elliptical constraint.

In summary, by applying the method of Lagrange multipliers, we have successfully determined the extreme values of the function xy within the specified ellipse.