Question

Use a spreadsheet to find the given extremum. In each case, assume that \(x, y\), and \(z\) are nonnegative. Maximize \(f(x, y, z)=x y z\) Constraints: \(x+3 y=6, x-2 z=0\)

Short answer

To find the maximum of the function \(f(x, y, z) = xyz\) under given constraints, we set up the Lagrange function, calculate its partial derivatives, solve the system of equations obtained, and then evaluate the function at these solutions.

Step-by-step solution

Setup Problem

Our problem is to maximize the function \(f(x, y, z) = xyz\), subject to the constraints \(x+3y=6\) and \(x-2z=0\). We can establish the Lagrange function like so: \(L(x, y, z, λ_1, λ_2) = xyz + λ_1(x + 3y - 6)+ λ_2(x-2z)\). The λs are Lagrange multipliers.

Find Partial Derivatives

Next, we calculate the first order partial derivatives of L with respect to x, y, z, λ_1, λ_2 and set each of them equal to zero. \(\frac{∂L}{∂x} = yz + λ_1 + λ_2 = 0\) \n\(\frac{∂L}{∂y} = xz + 3λ_1 = 0\) \n\(\frac{∂L}{∂z} = xy - 2λ_2 = 0\) \n\(\frac{∂L}{∂λ_1} = x+3y-6=0\) \n\(\frac{∂L}{∂λ_2} = x-2z=0\)

Solve the System of Equations

The goal now is to solve this system of equations. From the fourth and fifthe equation we can see they represent the original constraints on x, y, z. We can substitute x = 2z into the other equations to get equations we can solve. Solving this system can get complicated and is best done using a spreadsheet or a program.

Evaluate the Function

We evaluate the function f at the solutions obtained in step 3. This will give us the maximum value of the function under the given constraints, which is the solution to our problem.

Key concepts

Extremum Optimization

When we talk about extremum optimization, we are generally referring to the process of finding the maximum or minimum value of a function given some constraints. This is often a critical task in various fields such as economics, engineering, and physics where one needs to optimize a specific outcome. For example, a business might want to maximize profit while adhering to budgetary limits, or an engineer might need to minimize the weight of a structure without compromising its strength.

In the provided exercise, we're tasked with maximizing the function \(f(x, y, z) = xyz\) under two constraints: \(x+3y=6\) and \(x-2z=0\). This situation is perfectly suited for the method of Lagrange multipliers, which is a strategy used in calculus for finding local maxima and minima of functions subject to equality constraints. The method entails introducing additional variables, known as Lagrange multipliers (denoted by \(\lambda_i\)), and setting up a new function that incorporates the original function and the constraints.

The beauty of using Lagrange multipliers lies in its ability to convert a constrained problem into an unconstrained one. By doing so, the originally complex problem of extremum optimization becomes more manageable and can be systematically solved.

Partial Derivatives

Partial derivatives are the bread and butter of multivariable calculus. They are used to measure how a function changes as each variable is varied, while keeping all the other variables constant. In essence, partial derivatives give us the rate at which the function's value is changing in one direction, irrespective of the others.

With reference to our exercise, the function \(L(x, y, z, \lambda_1, \lambda_2)\) represents our Lagrange function where the partial derivatives with respect to \(x\), \(y\), \(z\), \(\lambda_1\), and \(\lambda_2\) are calculated. These derivatives are crucial as they help us understand how the function \(L\) behaves with small changes in each variable and are imperative in solving for the function's extrema. Setting these partial derivatives equal to zero allows us to find stationary points, which are potential candidates for optimization.

For instance, the partial derivative \(\frac{\partial L}{\partial x}\) gives information about how \(L\) changes when \(x\) is varied, holding \(y\), \(z\), \(\lambda_1\), and \(\lambda_2\) constant. It's a fundamental step in the method of Lagrange multipliers and assists in simplifying the process by breaking it down into more manageable pieces. In the exercise's context, finding the zeros of partial derivatives leads us to the values of \(x\), \(y\), \(z\), and the Lagrange multipliers that potentially maximize our original function \(f\).

System of Equations

A system of equations is a collection of two or more equations with a common set of variables. The goal is to find a solution that satisfies all equations simultaneously, which is often a point of intersection when visualized graphically. In many real-world problems, especially those involving constrained optimization, we use systems of equations to express relationships between variables and constraints.

In the step-by-step solution of our exercise, you will notice that after computing partial derivatives, we obtained a system of equations. This system combines both our constraints and the conditions for stationary points obtained from setting the partial derivatives to zero. Solving this system is vital as it yields the values of variables that could potentially optimize our function given the constraints.

To solve such a system, one might use techniques such as substitution, elimination, or even computational methods like matrix operations or numeric algorithms depending on the complexity. In our case, the fourth and fifth equations directly reflect the constraints on our variables, \(x\), \(y\), and \(z\), and we can use these to simplify the other equations and solve for the variables and Lagrange multipliers. Accurate and systematic resolution of this system is critical as it leads to the identification of the optimal values of the variables that satisfy our original optimization problem.