Problem 168 - Advanced Calculus

Question

a) Find the maxima and minima of \(\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{x}^{2}+\mathrm{y}^{2}\) on the ellipse \(\mathrm{G}(\mathrm{x}, \mathrm{y})=2 \mathrm{x}^{2}+3 \mathrm{y}^{2}=1\) b) Find the maxima value of \(\mathrm{f}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\mathrm{xyz}\) on the plane \((x / a)+(v / b)+(z / c)=1(a, b, c>0)\)

Step-by-Step Solution

Answer

The short answer for part a) is: the minimum value of f(x, y) is \(1/3\) at points \((0, 1/\sqrt{3})\) and \((0, -1/\sqrt{3})\), and the maximum value of f(x, y) is \(1/2\) at points \((1/\sqrt{2}, 0)\) and \((-1/\sqrt{2}, 0)\). For part b) the maximum value of f(x, y, z) occurs when x, y, and z satisfy the constraint \((x/a) + (y/b) + (z/c) = 1\). The exact maximum value depends on the values of a, b, and c.

Step 1: Set up the Lagrangian function

We can use the method of Lagrange multipliers to find the maxima and minima of a function subject to a constraint. First, set up the Lagrangian function L(x, y, λ) as follows: L(x, y, λ) = f(x, y) - λ(G(x, y) - 1) L(x, y, λ) = x^2 + y^2 - λ(2x^2 + 3y^2 - 1)

Step 2: Find the partial derivatives

Now we need to find the partial derivatives of L with respect to x, y, and λ, and set them equal to zero: ∂L/∂x = 2x - 4λx = 0 ∂L/∂y = 2y - 6λy = 0 ∂L/∂λ = 2x^2 + 3y^2 -1 = 0

Step 3: Solve the system of equations

Now we'll solve the system of equations given by the partial derivatives: 2x - 4λx = 0 → x(1 - 2λ) = 0 2y - 6λy = 0 → y(1 - 3λ) = 0 2x^2 + 3y^2 = 1 Using the first two equations, we have two possibilities: Case 1: x = 0, y(1 - 3λ) = 0 Case 2: y = 0, x(1 - 2λ) = 0 For Case 1: If x = 0, the third equation becomes 3y^2 = 1. Thus, we get y = ±1/√3, and λ = 1/3. For Case 2: If y = 0, the third equation becomes 2x^2 = 1. Thus, we get x = ±1/√2, and λ = 1/2.

Step 4: Obtain the critical points and evaluate f(x, y)

Now let's find the critical points and evaluate f(x, y) at these points: For Case 1: (0, 1/√3): f(0, 1/√3) = (0^2 + (1/√3)^2) = 1/3 (0, -1/√3): f(0, -1/√3) = (0^2 + (-1/√3)^2) = 1/3 For Case 2: (1/√2, 0): f(1/√2, 0) = ((1/√2)^2 + 0^2) = 1/2 (-1/√2, 0): f(-1/√2, 0) = ((-1/√2)^2 + 0^2) = 1/2

Step 5: Identify the maxima and minima

Finally, we can identify the maximum and minimum values of f(x, y) on the ellipse G(x, y): Minimum value: f(x, y) = 1/3 at points (0, 1/√3) and (0, -1/√3) Maximum value: f(x, y) = 1/2 at points (1/√2, 0) and (-1/√2, 0) b) To find the maximum value of f(x, y, z) = xyz on the plane (x/a) + (y/b) + (z/c) = 1, follow these steps:

Step 1: Set up the Lagrangian function

We can again use the method of Lagrange multipliers to find the maximum value of the function subject to a constraint. Set up the Lagrangian function L(x, y, z, λ) as follows: L(x, y, z, λ) = f(x, y, z) - λ((x/a) + (y/b) + (z/c) - 1) L(x, y, z, λ) = xyz - λ(x/a + y/b + z/c - 1)

Step 2: Find the partial derivatives

Now we need to find the partial derivatives of L with respect to x, y, z, and λ, and set them equal to zero: ∂L/∂x = yz - λ/a = 0 ∂L/∂y = xz - λ/b = 0 ∂L/∂z = xy - λ/c = 0 ∂L/∂λ = x/a + y/b + z/c - 1 = 0

Step 3: Solve the system of equations

Now we'll solve the system of equations given by the partial derivatives: yz = λ/a → λ = ayz xz = λ/b → λ = bxz xy = λ/c → λ = cxy Substitute the expressions for λ: ayz = bxz = cxy

Step 4: Obtain the critical points

Using the given condition (x/a) + (y/b) + (z/c) = 1, we can express one variable in terms of the others and plug this expression back into the expressions for λ: For example, let's express x in terms of y and z from the condition: x = a(1 - y/b - z/c) Now substitute this into the λ expressions: ayz = a^2(yz - yy/b - yz/c) = cxy Cancel out the y terms and divide by ac: z = x(1 - 1/b - z/c) Now isolate z: z(b+1) = x(b-c) z = x(b-c)/(b+1) Now substitute z back into the condition (x/a) + (y/b) + (z/c) = 1, we have: x/a + y/b + x(b-c)/((b+1)c) = 1

Step 5: Find the maximum value

Now we have the equation for f(x,y,z) = xyz subject to the constraint on a simplified form. The optimization problem involves maximizing the product of three variables under this constraint. The maximum occurs when x, y, and z are equal. Solve for x, y, and z, and then find the maximum value of the function. In this particular problem, the exact maximum value will depend on the parameters a, b, and c. The maximum value of f(x, y, z) = xyz on the plane (x/a) + (y/b) + (z/c) = 1 will occur when x, y, and z are set such that the constraint equation is satisfied.

Key Concepts

Maxima and Minima

In mathematics, finding the **maxima** and **minima** of a function involves determining the highest and lowest values a function can achieve within a given domain or under certain constraints. The maxima and minima, also known as **extrema**, are critical for understanding the behavior of functions in optimization problems. These points occur where the partial derivatives of the function equal zero, indicating potential peaks or troughs.
The search for extrema is a common goal in mathematics, particularly in optimization problems where specific conditions or constraints such as the equation of a curve or surface are applied. In our exercise, the function to be optimized is subject to an ellipse constraint, which is a common scenario in applied mathematics and economics.

Ellipse Constraint

An **ellipse constraint** refers to a restriction applied to the variables of a function, molding the feasible region to an elliptical shape. This constraint ensures that the solution to an optimization problem lies on the boundary of the ellipse. In the problem, the constraint is given by the equation \(2x^2 + 3y^2 = 1\), forming an ellipse in the xy-plane.
The ellipse boundary guides where the function can be optimized, limiting potential solutions to points on the ellipse. Such a setting is typical in problems involving multivariable functions and constraints. The elliptical shape reflects a balance between the variables, ensuring neither exceeds certain bounds, thereby modeling real-life scenarios such as physical limits or resources.

Partial Derivatives

**Partial derivatives** play a crucial role in analyzing functions of multiple variables. They measure how the function changes as one particular variable is varied while keeping others constant. In optimization, they help identify critical points by setting the partial derivatives of the function equal to zero.

• The notation \( \frac{\partial f}{\partial x} \) represents the partial derivative of \( f(x, y) \) with respect to \( x \).
• In the given problem, finding the partial derivatives of the Lagrangian function helps derive conditions for both x and y that are necessary for identifying potential maxima and minima.

By understanding these derivatives, one can pinpoint where the function's slope equals zero, thus identifying critical points on the constraint boundary — a crucial step in solving optimization problems.

Optimization Problems

**Optimization problems** involve finding the best solution based on given constraints and conditions. They encompass finding maxima, minima, or optimal values of a function under specific scenarios. These problems are pivotal in fields like economics, engineering, and logistics.
In the exercise, optimization is performed using the method of **Lagrange multipliers**, a powerful technique that incorporates constraints directly into the function using an auxiliary variable called the Lagrange multiplier. This method:

• Transforms a constrained optimization problem into a system of equations.
• Allows solving for critical points where the optimal values occur.

This technique ensures that the constraint's impact on the solution is fully accounted for, leading to the best possible results under the specified conditions, such as an ellipse or plane in multidimensional space.