Question

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

Short answer

The maximum of the function \(f(x, y, z)=x+y+z\) under the constraint \(x^{2}+y^{2}+z^{2}=1\) is \(\sqrt{3}\). In detail, this maximum is reached at the point \((x, y, z) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\)

Step-by-step solution

Form The Lagrange

Firstly, form the Lagrange. Let's call the Lagrangian function \(L(x, y, z, \lambda)\) which is the sum of original function \(f(x, y, z)\) and the product of a new variable \(\lambda\) (the Lagrange multiplier) with the constraint: \(L(x, y, z, \lambda) = f(x, y, z) - \lambda(g(x, y, z) - c)\)We substitute \(f(x, y, z)\) and \(g(x, y, z)\) with the given function and constraint. Now, it becomes:\(L(x, y, z, \lambda) = x + y + z - \lambda(x^{2} + y^{2} + z^{2} - 1)\)

Find The Gradient of The Lagrangian

Find the gradient of the Lagrangian (partial derivatives equal to zero): \(\nabla L = 0 \rightarrow \frac{\partial L}{\partial x} = 1 - 2\lambda x = 0\),\(\frac{\partial L}{\partial y} = 1 - 2\lambda y = 0\),\(\frac{\partial L}{\partial z} = 1 - 2\lambda z = 0\),Also, the derivative with respect to \(\lambda\) gives back our original constraint. After calculating, we will have values of \(x, y, z\) which are \(x= y= z= \frac{1}{\sqrt{3}}\) and \(\lambda = \sqrt{3}\).

Check The Second Derivative

Now, substitute these found values into the second order partial derivatives of \(L\) for ensuring that a maximum point is reached: \(\frac{\partial^{2} L}{\partial x^{2}} = -2\lambda\), \(\frac{\partial^{2} L}{\partial y^{2}} = -2\lambda\), \(\frac{\partial^{2} L}{\partial z^{2}} = -2\lambda\), After substitution, we got \(negative\) values for all which indicates a maximum value.

Substitute Solution into Original Function

Now substitute the found \(x, y, z\) values into the \(f(x, y, z)\) function. We get \(f(x, y, z)= \frac{3}{\sqrt{3}}= \sqrt{3}\)

Key concepts

Constrained Optimization

Constrained optimization is a powerful technique used to find maximum or minimum values of a function under certain restrictions.
It helps us ensure that we not only optimize our objective but also adhere to certain limitations or bounds represented by constraints.
In our given exercise, the function to be maximized is \( f(x, y, z) = x + y + z \), and the constraint is \( x^2 + y^2 + z^2 = 1 \).

• Objective Function: The main function you are trying to maximize or minimize.
• Constraint: A condition that the solution must satisfy. Here it's \( x^2 + y^2 + z^2 = 1 \).
• Lagrange Multipliers: These are used to solve constrained optimization problems by transforming them into unconstrained problems.

The Lagrange multiplier method involves creating a new function by combining the objective and the constraint functions.
This is done using a multiplier, \( \lambda \), which balances the trade-off between the function we want to optimize and the constraint conditions.
In this way, we change the constrained problem into a problem of finding extreme values of this new function.

Partial Derivatives

Partial derivatives play a crucial role in the process of constrained optimization.
By taking the partial derivatives of the Lagrangian, we find the rate at which the function changes with respect to one variable while keeping others constant.
In our specific example, after forming the Lagrangian \( L(x, y, z, \lambda) = x + y + z - \lambda(x^2 + y^2 + z^2 - 1) \),

• We derive the partial derivatives with respect to \( x \), \( y \), and \( z \).
• Setting these partial derivatives equal to zero \( abla L = 0 \) helps us to find critical points where the extremum might occur.

These steps allow us to solve for \( x, y, z, \) and \( \lambda \), providing the values needed to check if a maximum or minimum has been achieved.
In the solution, the results are \( x = y = z = \frac{1}{\sqrt{3}} \), and \( \lambda = \sqrt{3} \).
This is a fundamental step in verifying that we have reached an actual extremum point.

Extreme Values

In optimization problems, finding extreme values refers to identifying either the maximum or minimum value of a function.
With Lagrange multipliers, once the critical points have been determined, the next step involves checking whether these points are indeed extrema.
By calculating the second order partial derivatives of the Lagrangian, we confirm the nature of these critical points.

• For our Lagrangian, the second derivatives were \( \frac{\partial^2 L}{\partial x^2} = -2\lambda \), and similarly for \( y \) and \( z \).
• Simplifying this with our found \( \lambda \) value, we get negative numbers, which indicate the presence of a maximum.

Finally, substituting these values back into the original function confirms that the function achieves its maximum under the given conditions.
The computed maximum value \( \sqrt{3} \) eloquently meets the constraint \( x^2 + y^2 + z^2 = 1 \), thus effectively solving the constrained optimization problem.