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