Problem 18 - Calculus an Applied Approach

Question

Use Lagrange multipliers to find the given extremum. In each case, assume that \(x, y,\) and \(z\) are positive. $$ \begin{array}{l}{\text { Maximize } f(x, y, z)=x^{2} y^{2} z^{2}} \\ {\text { Constraint: } x^{2}+y^{2}+z^{2}=1}\end{array} $$

Step-by-Step Solution

Answer

The maximum value of the function \(f(x, y, z)=x^2 y^2 z^2\) subject to the constraint \(x^2 + y^2 + z^2 = 1\) is \(f(x, y, z) = \frac{1}{27}\) at the point \((x,y,z)=(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})\).

Step 1: Setting up the Lagrange function

The Lagrange function is \(L(x, y, z, \lambda) = f(x, y, z) - \lambda g(x, y, z)\), where \(\lambda\) is the Lagrange multiplier. Substituting \(f\) and \(g\) gives: \[L(x, y, z, \lambda)=x^2 y^2 z^2 - \lambda (x^2 + y^2 + z^2 - 1)\]

Step 2: Computing the Gradient

Set the gradients of \(L\) equal to zero. In other words, find the partial derivatives with respect to \(x, y, z, \lambda\), and set them equal to 0: \[\frac{\partial L}{\partial x} = 0, \frac{\partial L}{\partial y} = 0, \frac{\partial L}{\partial z} = 0, \frac{\partial L}{\partial \lambda}=0\]

Step 3: Solving the System of Equations

We will now solve the equations obtained in step 2. The system of equations would be: \[2xyz^2 - 2\lambda x = 0\] \[2x^2 yz^2 - 2\lambda y = 0\] \[2x^2 y^2 z - 2\lambda z = 0\] \[x^2 + y^2 + z^2 - 1 = 0\] From the first three equations we can derive that \(x=y=z\) and \(\lambda = xyz\). Substituting these into the fourth equation will give the solutions for \(x,y,z,\lambda\).

Step 4: Finding all Solutions

Upon solving the system of equations as above we obtain two solutions, \((x,y,z)=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\) where \(\lambda = \frac{1}{\sqrt{3}}\) and \((x,y,z)=\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)\) where \(\lambda = -\frac{1}{\sqrt{3}}\).

Step 5: Identifying the Maxima

Substitute the solutions into \(f(x, y, z)\). The corresponding \(f\) values will tell if it is a maximum, minimum, or saddle point. Here the maximum is obtained at \((x,y,z)=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\) with the maximum value being \(f(x, y, z)= \frac{1}{27}\).

Key Concepts

Extremum

An extremum refers to the points at which a function reaches its minimum or maximum value. In the context of the Lagrange multipliers method, we are often trying to find these extremum points under a given constraint. For example, in the provided problem, we want to find the maximum value of the function \( f(x, y, z) = x^2 y^2 z^2 \) under the constraint \( x^2 + y^2 + z^2 = 1 \). This means we are looking for the point where the function hits its peak value, given that these points must satisfy the constraint. By using Lagrange multipliers, we can efficiently locate these points without directly solving complicated systems which incorporate the constraint.

Gradient

The gradient is a vector that points in the direction of the greatest rate of increase of a function. It is a crucial element when using Lagrange multipliers because it helps in setting up the necessary conditions for finding extremum points. In our scenario, we calculate the gradient of the Lagrange function \( L(x, y, z, \lambda) = f(x, y, z) - \lambda g(x, y, z) \) and set it to zero:

• \( \frac{\partial L}{\partial x} = 0 \)
• \( \frac{\partial L}{\partial y} = 0 \)
• \( \frac{\partial L}{\partial z} = 0 \)
• \( \frac{\partial L}{\partial \lambda} = 0 \)

These equations ensure that the gradients of the function and the constraint are parallel, which is a fundamental criterion when looking for the extremum under constraints.

System of Equations

A system of equations is a set of equations that are solved together, as they share several variables. When employing Lagrange multipliers, we often end up with a system derived from setting the gradient of the Lagrange function to zero. In this particular example, the system of equations is:

• \( 2xyz^2 - 2\lambda x = 0 \)
• \( 2x^2 yz^2 - 2\lambda y = 0 \)
• \( 2x^2 y^2 z - 2\lambda z = 0 \)
• \( x^2 + y^2 + z^2 - 1 = 0 \)

Solving this system allows us to find values of \( x, y, z, \) and \( \lambda \) that satisfy both the function and the constraint, ultimately helping in finding extremums.

Partial Derivatives

Partial derivatives measure how a function changes as we alter one of its variables, holding the others constant. They are critical when setting up the gradient in the method of Lagrange multipliers because they provide us with the necessary conditions to establish extremum points. In the Lagrange function, by taking partial derivatives with respect to each variable \( x, y, z, \text{and } \lambda \), we set them equal to zero to ensure all gradients involved are parallel. This means:

• \( \frac{\partial L}{\partial x} \) evaluates the change in the Lagrange function if we only change \( x \).
• Analogously, \( \frac{\partial L}{\partial y} \) and \( \frac{\partial L}{\partial z} \) do the same for \( y \) and \( z \), respectively.
• Finally, \( \frac{\partial L}{\partial \lambda} \) corresponds to the constraint, ensuring that we correctly balance the original function and the constraint equation.

Mastering partial derivatives is essential for navigating through any multivariable calculus problem involving constraints.