Question

Find the maximum of \(f(x, y)=4 x^{2}-4 x y+y^{2}\) subject to the constraint \(x^{2}+y^{2}=1\).

Short answer

The maximum value is 4 at \( (x, y) = (1, 0) \).

Step-by-step solution

Define the Objective and Constraint Functions

Identify the given function to optimize, which is the objective function, and the restriction it is subject to. The objective function is given by:\[ f(x, y) = 4x^2 - 4xy + y^2 \]and the constraint is:\[ g(x, y) = x^2 + y^2 = 1. \]

Calculate the Gradient of the Functions

Find the gradients of the objective function and the constraint:1. The gradient of \(f\): \[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (8x - 4y, -4x + 2y). \]2. The gradient of \(g\): \[ abla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) = (2x, 2y). \]

Apply the Method of Lagrange Multipliers

Introduce a Lagrange multiplier, \(\lambda\), and set up the system of equations by equating the gradients of the objective function and the constraint times the multiplier:\[ abla f = \lambda abla g. \]This gives the following system:1. \(8x - 4y = 2\lambda x\)2. \(-4x + 2y = 2\lambda y\)

Solve the System of Equations

Solve the system of equations obtained from the application of Lagrange multipliers:1. From \(8x - 4y = 2\lambda x\), we get \(2x(4 - \lambda) = 4y\).2. From \(-4x + 2y = 2\lambda y\), we get \(-4x = 2y(\lambda - 1)\).Substitute \(x^2 + y^2 = 1\) to solve these equations together.

Analyze the Critical Points

Determine critical points from the system of equations and verify them under constraint:- One solution is \(x = 1, y = 0\), which satisfies \(x^2 + y^2 = 1\).- Substitute back into \(f(x, y)\) to find \(f(1, 0) = 4.\)- Find other solutions and evaluate potential maxima by substituting them into \(f(x, y)\).

Evaluate the Known Values

Since there might be multiple points, evaluate at those and compare: Testing other potential points like \((x, y) = (0, 1)\) which also fit the constraint, evaluating function there yields \(-1.\)The maximum value from evaluated solutions is determined.

Confirm the Maximum Value

Check values to conclude regarding the optimal value. As evaluated, maximum occurs clearly at \((x, y) = (1, 0)\) with \(f(x, y) = 4.\)

Key concepts

Constrained Optimization

When solving optimization problems where there are limitations or restrictions, we deal with **constrained optimization**. This means we are trying to find the maximum or minimum value of a function within a set of constraints. These constraints are often equations or inequalities that restrict the values that we can assign to the variables of our function.

In our exercise, we aim to maximize the function \[f(x, y) = 4x^2 - 4xy + y^2\]subject to the constraint \[g(x, y) = x^2 + y^2 = 1.\]

Such constraints can represent conditions in the real world, like resources or gaps that cannot be exceeded, and are crucial in ensuring that potential solutions are practical. In this problem, the constraint is the equation of a circle, which means our solution is confined to the circumference of this circle.

Gradient

The **gradient** is a vector that stores all partial derivatives of a multivariable function. For a function defined as \[f(x, y)\],its gradient, \(abla f\),is a vector with components corresponding to the rate of change along each axis, \[(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}).\]

In the context of optimization, particularly constrained optimization, the gradient tells us the direction of steepest ascent of the function. In our problem, to find where this direction aligns with the constraint, we equate the gradient of the objective function to that of the constraint multiplied by a scalar.

This scalar, known as the Lagrange multiplier, allows us to explore the surface of the constraint, ensuring the best possible value under the given condition.

Critical Points

**Critical points** are where the first derivatives of a function (partial derivatives) are zero or undefined. These points are significant candidates for potential maxima or minima of the function.

In constrained optimization, critical points can be found using the method of Lagrange multipliers. We find where the gradient of our function and the constraint align after setting them equal with the introduction of a Lagrange multiplier, \(\lambda\).

In practice, this provides a system of equations that we solve to find these critical points. These points must lie on the constraint curve in our example because they adhere to the constraint that \(x^2 + y^2 = 1.\)

By analyzing each critical point, we can evaluate the function values to determine whether they are maxima, minima, or saddle points.

Calculus

**Calculus** is essential in studying and solving optimization problems through its focus on changes and motion. In this exercise, we primarily use differential calculus, which involves derivatives and gradients.

The derivative gives us the rate of change, which helps identify increases or decreases in the function's value. Moreover, understanding second derivatives can help distinguish whether critical points are maxima, minima, or saddle points.

Through the method of Lagrange multipliers, calculus helps solve the constraint optimization problem by introducing a multiplier that balances both the objective function and the constraint.

By applying these concepts, calculus enables us to elegantly navigate the complex landscape of functions bound by constraints and zero in on optimal solutions.