Problem 167 - Advanced Calculus

Question

Find the maximum of \(\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{xy}\) on the curve $$ G(x, y)=(x+1)^{2}+y^{2}=1 $$ assuming that such a maximum exists.

Step-by-Step Solution

Answer

The maximum value of \(f(x, y) = xy\) on the curve \((x+1)^2 + y^2 = 1\) is \(\frac{2\sqrt{2}}{9}\), which occurs at the point \((\frac{-1}{3}, -\frac{2\sqrt{2}}{3})\).

Step 1: Set up the system of equations using Lagrange multipliers

Introduce a new variable, \(\lambda\), called the Lagrange multiplier. Write down the system of equations: \[ \begin{aligned} \nabla f(x, y) &= \lambda \nabla G(x, y) \\ G(x, y) &= 1 \end{aligned} \] where \(\nabla\) denotes the gradient. Compute the gradients of \(f(x, y)\) and \(G(x, y)\): \[ \begin{aligned} \nabla f(x, y) &= \begin{bmatrix} y \\ x \end{bmatrix} \\ \nabla G(x, y) &= \begin{bmatrix} 2(x+1) \\ 2y \end{bmatrix} \end{aligned} \] Substitute the gradients into the system of equations: \[ \begin{aligned} \begin{bmatrix} y \\ x \end{bmatrix} &= \lambda \begin{bmatrix} 2(x+1) \\ 2y \end{bmatrix} \\ (x+1)^2 + y^2 &= 1 \end{aligned} \]

Step 2: Solve the system of equations

From the first equation in the system, we get two equations: \[ \begin{aligned} y &= 2\lambda (x+1) \\ x &= 2\lambda y \end{aligned} \] Substitute the first equation into the second equation: \[x = 2\lambda (2\lambda (x+1))\] This is a quadratic equation in the variable \(x\). Solve for \(x\): \[x = \frac{-1}{3}\] or \(x = -1\) For each value of \(x\), we can find the corresponding value of \(y\) by using the constraint equation \(G(x, y) = 1\). When \(x = \frac{-1}{3}\), the constraint equation implies \((\frac{-1}{3} + 1)^2 + y^2 = 1\). Solve for \(y\): \[y = \pm \frac{2\sqrt{2}}{3}\] When \(x=-1\), the constraint equation implies \((-1 + 1)^2 + y^2 = 1\). Solve for \(y\): \[y = 0\] Thus, we have three critical points: \((\frac{-1}{3}, \frac{2\sqrt{2}}{3})\), \((\frac{-1}{3}, -\frac{2\sqrt{2}}{3})\), and \((-1, 0)\).

Step 3: Compute the function values at the critical points

Compute the values of \(f(x, y) = xy\) at the three critical points: 1. When \((x, y) = (\frac{-1}{3}, \frac{2\sqrt{2}}{3})\), \(f(x, y) = (\frac{-1}{3})(\frac{2\sqrt{2}}{3}) = -\frac{2\sqrt{2}}{9}\). 2. When \((x, y) = (\frac{-1}{3}, -\frac{2\sqrt{2}}{3})\), \(f(x, y) = (\frac{-1}{3})(-\frac{2\sqrt{2}}{3}) = \frac{2\sqrt{2}}{9}\). 3. When \((x, y) = (-1, 0)\), \(f(x, y) = (-1)(0) = 0\).

Step 4: Determine the maximum

The maximum value of \(f(x, y) = xy\) on the curve \((x+1)^2 + y^2 = 1\) is \(\frac{2\sqrt{2}}{9}\), which occurs at the point \((\frac{-1}{3}, -\frac{2\sqrt{2}}{3})\).

Key Concepts

Gradient Calculation

In calculus, the gradient is an essential tool that tells us how a function changes at any given point. The gradient consists of partial derivatives and is denoted by \(abla\). It points in the direction of the steepest slope. For a function \(f(x, y)\), it is calculated as:

• \(abla f(x, y) = \left[ \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right]\)

Understanding the gradient is crucial when using the method of Lagrange multipliers. This method involves computing gradients of both the objective function and the constraint. In our example, we calculate the gradients:

• \(abla f(x, y) = \begin{bmatrix} y \ x \end{bmatrix}\)
• \(abla G(x, y) = \begin{bmatrix} 2(x+1) \ 2y \end{bmatrix}\)

This calculation sets the stage for finding points where the function constrained by \(G(x, y)\) is optimized. The beauty of gradient calculation lies in its ability to simplify complex multivariable problems into manageable parts, revealing where the crucial changes occur.

Constraint Optimization

Constraint optimization tackles problems where you want to optimize a function subject to certain conditions. This is where Lagrange multipliers shine. They enable finding extrema (either maximum or minimum points) of a function subject to constraints.
In this context, the classical approach begins by defining a new variable, the Lagrange multiplier \(\lambda\). The goal is to solve:

• \(abla f(x, y) = \lambda abla G(x, y)\)
• \(G(x, y) = 1\)

The above system of equations ensures the optimization of \(f(x, y)\) while respecting the constraint \(G(x, y) = 1\). It's like finding safe paths on a map, where you can't go beyond the lines drawn by \(G(x, y)\).
By aligning derivatives (gradients) with Lagrange multipliers, we can gain insights into critical points where the objective function could potentially reach its maximum or minimum under a given constraint.

Critical Points

Critical points are where a function's behavior changes, marking potential locations for maxima, minima, or saddle points. In optimization, especially with constraints, these are the solutions to Lagrange's equations.
The process involves substituting the gradients and solving for critical points. In our scenario, solving:

• \(y = 2\lambda (x+1)\)
• \(x = 2\lambda y\)

leads to specific values of \((x, y)\) that satisfy both the function and the constraint.
By substituting calculated coordinates back into the constraint equation \((x+1)^2 + y^2 = 1\), we're reaffirming their legitimacy as plotted points on the constraint curve.
Analyzing these critical points, such as \((\frac{-1}{3}, \frac{2\sqrt{2}}{3})\), helps determine where an objective function like \(xy\) can reach its highest or lowest value. These critical spots are fundamental in identifying the best outcome within the defined boundaries.