Problem 65 - Calculus Early Transcendentals

Question

Let \(x, y,\) and \(z\) be nonnegative numbers with \(x+y+z=200\). a. Find the values of \(x, y,\) and \(z\) that minimize \(x^{2}+y^{2}+z^{2}\). b. Find the values of \(x, y,\) and \(z\) that minimize \(\sqrt{x^{2}+y^{2}+z^{2}}\). c. Find the values of \(x, y,\) and \(z\) that maximize \(x y z\). d. Find the values of \(x, y,\) and \(z\) that maximize \(x^{2} y^{2} z^{2}\).

Step-by-Step Solution

Answer

Answer: a. The minimum value of \(x^2+y^2+z^2\) occurs at \(x = y = z = \frac{200}{3}\). b. The minimum value of \(\sqrt{x^2+y^2+z^2}\) occurs at \(x = y = z = \frac{200}{3}\). c. The maximum value of \(xyz\) occurs at \(x = y = z = \frac{200}{3}\). d. The maximum value of \(x^2 y^2 z^2\) occurs at \(x = y = z = \frac{200}{3}\).

Step 1: Setting up the Lagrangian function

Set up the Lagrangian function for minimizing \(f(x, y, z) = x^2+y^2+z^2\) subject to the constraint \(g(x, y, z) = x + y + z - 200=0\). The Lagrangian function is: $$L(x, y, z, \lambda) = x^2+y^2+z^2 + \lambda(x+y+z-200).$$

Step 2: Taking partial derivatives

Compute the partial derivatives of \(L\) with respect to \(x, y, z,\) and \(\lambda\): $$\begin{aligned} \frac{\partial L}{\partial x} &= 2x + \lambda, \\ \frac{\partial L}{\partial y} &= 2y + \lambda, \\ \frac{\partial L}{\partial z} &= 2z + \lambda, \\ \frac{\partial L}{\partial \lambda} &= x+y+z-200. \end{aligned}$$

Step 3: Solving the system of equations

Set the partial derivatives equal to 0 and solve the system of equations: $$\begin{aligned} 2x + \lambda &= 0, \\ 2y + \lambda &= 0, \\ 2z + \lambda &= 0, \\ x+y+z-200 &= 0. \end{aligned}$$ From the first three equations, we get \(x = y = z\), and from the last equation, we get \(x + y + z = 200\).

Step 4: Finding the values of x, y, and z

Since we have \(x=y=z\), we plug this into the constraint equation to get \(3x=200\), which results in \(x=y=z=\frac{200}{3}\). Therefore, the values of \(x, y,\) and \(z\) that minimize \(x^2+y^2+z^2\) are \(x = y = z = \frac{200}{3}\). b. Minimize \(\sqrt{x^{2}+y^{2}+z^{2}}\)

Step 5: Observation

Note that we have already minimized \(x^{2}+y^{2}+z^{2}\) in part (a). Since the square root function is strictly increasing, it preserves the order of the values. This means that the minimum value for \(\sqrt{x^{2}+y^{2}+z^{2}}\) occurs at the same point as the minimum value for \(x^{2}+y^{2}+z^{2}\).

Step 6: Minimizing the square root expression

The minimum value of \(\sqrt{x^{2}+y^{2}+z^{2}}\) occurs at the same point as the minimum value of \(x^2+y^2+z^2\), which we found in part (a) to be \(x = y = z = \frac{200}{3}\). c. Maximize \(xyz\)

Step 1: Setting up the Lagrangian function

Set up the Lagrangian function for maximizing \(f(x, y, z) = xyz\) subject to the constraint \(g(x, y, z) = x + y + z - 200=0\). The Lagrangian function is: $$L(x, y, z, \lambda) = xyz + \lambda(x+y+z-200).$$

Step 2: Taking partial derivatives

Compute the partial derivatives of \(L\) with respect to \(x, y, z,\) and \(\lambda\): $$\begin{aligned} \frac{\partial L}{\partial x} &= yz + \lambda, \\ \frac{\partial L}{\partial y} &= xz + \lambda, \\ \frac{\partial L}{\partial z} &= xy + \lambda, \\ \frac{\partial L}{\partial \lambda} &= x+y+z-200. \end{aligned}$$

Step 3: Solving the system of equations

Set the partial derivatives equal to 0 and solve the system of equations: $$\begin{aligned} yz + \lambda &= 0, \\ xz + \lambda &= 0, \\ xy + \lambda &= 0, \\ x+y+z-200 &= 0. \end{aligned}$$ From the first three equations, we get \(yz = xz = xy\), and from the last equation, we get \(x + y + z = 200\).

Step 4: Finding the values of x, y, and z

Since \(yz = xz = xy\), we have \(x=y=z\). Plugging this into the constraint equation, we get \(3x=200\), which leads to \(x=y=z=\frac{200}{3}\). Therefore, the values of \(x, y,\) and \(z\) that maximize \(xyz\) are \(x = y = z = \frac{200}{3}\). d. Maximize \(x^{2} y^{2} z^{2}\)

Step 1: Setting up the Lagrangian function

Set up the Lagrangian function for maximizing \(f(x, y, z) = x^2 y^2 z^2\) subject to the constraint \(g(x, y, z) = x + y + z - 200=0\). The Lagrangian function is: $$L(x, y, z, \lambda) = x^2 y^2 z^2 + \lambda(x+y+z-200).$$

Step 2: Taking partial derivatives

Compute the partial derivatives of \(L\) with respect to \(x, y, z,\) and \(\lambda\): $$\begin{aligned} \frac{\partial L}{\partial x} &= 2xy^2z^2 + \lambda, \\ \frac{\partial L}{\partial y} &= 2x^2yz^2 + \lambda, \\ \frac{\partial L}{\partial z} &= 2x^2y^2z + \lambda, \\ \frac{\partial L}{\partial \lambda} &= x+y+z-200. \end{aligned}$$

Step 3: Solving the system of equations

Set the partial derivatives equal to 0 and solve the system of equations: $$\begin{aligned} 2xy^2z^2 + \lambda &= 0, \\ 2x^2yz^2 + \lambda &= 0, \\ 2x^2y^2z + \lambda &= 0, \\ x+y+z-200 &= 0. \end{aligned}$$ From the first three equations, we get \(2xy^2z^2 = 2x^2yz^2 = 2x^2y^2z\), and from the last equation, we get \(x + y + z = 200\).

Step 4: Finding the values of x, y, and z

Since \(2xy^2z^2 = 2x^2yz^2 = 2x^2y^2z\), we have \(x=y=z\). Plugging this into the constraint equation, we get \(3x=200\), which leads to \(x=y=z=\frac{200}{3}\). Therefore, the values of \(x, y,\) and \(z\) that maximize \(x^2 y^2 z^2\) are \(x = y = z = \frac{200}{3}\).

Key Concepts

Optimization in Mathematics

Optimization is all about finding the best possible solution or outcome in a given situation. In mathematics, this means either maximizing or minimizing a certain function. Imagine you have a garden and you want to plant as many flowers as possible—your goal is to maximize the number of flowers. Optimization works similarly, using mathematical strategies to find the highest or lowest possible value of a function.

In problems involving optimization, you often work with functions that describe the situation. You look for the "optimal" values that will give you a peak (maximum) or a trough (minimum). If we optimize the function itself, we might use techniques like calculus and algebra to reach our solution. Understanding optimization helps you make better decisions on how best to use resources or achieve certain goals.

Understanding Partial Derivatives

When dealing with functions of multiple variables, such as \(x, y, \) and \(z\), partial derivatives become very useful. They help us understand how changes in one variable affect the function’s value while keeping the others constant. It’s like seeing how changing one ingredient affects the taste of a recipe while all other ingredients stay the same.

In the context of Lagrange multipliers, you derive partial derivatives of a function to understand how each variable contributes to the overall value. This allows you to set equations that help find critical points, which can lead to finding maxima or minima. By calculating \(\frac{\partial L}{\partial x}\), \(\frac{\partial L}{\partial y}\), and \(\frac{\partial L}{\partial z}\), you can decipher each variable's role in achieving the goal of the optimization problem.

• Partial derivatives are denoted as \(\frac{\partial}{\partial x}\), for example, which shows the rate of change of the function with respect to x.
• They are critical in identifying critical points where maxima or minima are potential.
• Setting these partial derivatives to zero helps establish conditions for optimization.

Constrained Optimization with Lagrange Multipliers

Constrained optimization involves finding an optimal solution to a problem that must satisfy certain restrictions or "constraints." Imagine you want to create the largest possible garden (maximize area) but are limited by the size of your backyard. In mathematical terms, these are constraints.

Lagrange multipliers come into play when you have such constraints. It's a method that helps find the maximum or minimum of a function subject to constraints. The basic idea is that you consider both the function you want to optimize and the constraint together using a Lagrange function. Its structure is \(L(f, g, \lambda) = f(x, y, z) + \lambda \, g(x, y, z)\).

Here’s how it works in steps:

• Set up the Lagrangian by combining your function and the constraint with a multiplier, \(\lambda\).
• Take partial derivatives of this combined function with respect to all variables, including \(\lambda\).
• Set these derivatives equal to zero to find the critical points.
• Solve this system of equations to find the values that optimize the function while satisfying the constraint.

This powerful method allows you to navigate complex optimization problems where direct solutions might not be straightforward.