91Ó°ÊÓ

Constraint optimization is a powerful method used in calculus and applied mathematics to find the maximum or minimum of a function subject to some constraints. In this exercise, we are looking to minimize or maximize the function \(f(x, y) = x^2 + y^2\) under a given constraint \(g(x, y) = 2x^2 + 3xy + 2y^2 - 7 = 0\).
The key idea is to use a technique called the method of Lagrange multipliers, which helps tackle problems where direct optimization is tricky because of the constraints. Here, instead of only optimizing the original function \(f(x, y)\), we construct a new function by adding the constraint into the optimization as a modifier, with the help of a Lagrange multiplier \(\lambda\).
This allows us to simultaneously solve both the optimization requirement and satisfy the constraint. The Lagrangian function thus formed helps convert the problem into a system we can solve using partial derivatives.

• Lagrangian function: \(L(x, y, \lambda) = x^2 + y^2 - \lambda(2x^2 + 3xy + 2y^2 - 7)\)
• Our goal is to find values of \(x, y, \lambda\) that minimize or maximize \(f(x, y)\).

Partial derivatives are essential tools in multivariable calculus used to compute the derivative of a function with respect to one variable while keeping others constant. They help us understand how a function changes when its inputs change.
In this problem, the Lagrangian \(L(x, y, \lambda)\) is differentiated partially with respect to each variable \(x, y, \) and \(\lambda\). This step is critical as it helps us form a system of equations, which we then solve for the optimization process.
The partial derivatives of the Lagrangian are:

• With respect to \(x\): \(\frac{\partial L}{\partial x} = 2x - \lambda(4x + 3y)\)
• With respect to \(y\): \(\frac{\partial L}{\partial y} = 2y - \lambda(3x + 4y)\)
• With respect to \(\lambda\): \(\frac{\partial L}{\partial \lambda} = -(2x^2 + 3xy + 2y^2 - 7)\)

These derivatives give us insight into how to adjust each variable to maintain optimal conditions under the constraint. The goal of taking these partial derivatives and setting them to zero is to find the stationary points where the function does not increase or decrease, indicating potential extrema.

After computing the partial derivatives of the Lagrangian, we set them equal to zero to form a system of equations. Solving these will yield the values of \(x\), \(y\), and \(\lambda\) that satisfy the constraints and minimize or maximize our target function.
The system of equations generated is:

• \(2x - \lambda(4x + 3y) = 0\)
• \(2y - \lambda(3x + 4y) = 0\)
• \(2x^2 + 3xy + 2y^2 - 7 = 0\)

These equations come from setting the partial derivatives of the Lagrangian to zero. The solutions to this set of equations, known as the critical points, are candidates for the minima or maxima of \(f(x, y)\).
Once the values for \(x\) and \(y\) are obtained, they can be plugged back into the original function \(f(x, y)\) to find the optimized extreme values. This systematic approach allows us to handle optimization problems with constraints efficiently, leveraging the power of calculus.