91Ó°ÊÓ

The substitution method is a powerful technique in calculus used to simplify complex optimization problems by reducing the number of variables. It involves substituting one variable with another or with an expression involving other variables to make the equation easier to manage.

In our example, the goal is to maximize the function f(x, y, z) given the constraint 3x = y. By substituting the constraint into the function, we eliminate the variable y and rewrite f in terms of x and z only. This simplification allows us to focus on finding the critical points by working with a function of fewer variables.

The substitution in our problem is straightforward: y gets replaced by 3x, reducing the original problem into a two-variable optimization problem. It's essential to perform the substitution correctly to avoid any errors that could lead to incorrect critical points and, subsequently, an incorrect solution.

When dealing with functions of multiple variables, as is the case with f(x, y, z), finding the rate of change with respect to one variable while keeping others constant is done through partial differentiation. This technique is critical in identifying the critical points needed to solve optimization problems.

After substitution, we partially differentiated the new function f(x, z) with respect to both x and z independently. The expressions \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial z}\) represent the slopes of the function in the direction of x and z, respectively.

To find critical points, we set these partial derivatives to zero because at maximum and minimum points of a function, the slope is zero. By solving these equations, we can pinpoint the exact location of these critical points on the function's surface.

In calculus, critical points are where a function's derivative is zero or undefined, indicating potential maximums, minimums, or saddle points. To solve optimization problems, identifying the critical points is crucial as they pinpoint where the function peaks or dips.

In our step by step solution, after substitution and partial differentiation, we set the derivatives equal to zero to find x and z values that give us critical points. These values are x = 1/10 and z = 1/2 – which we then used to calculate the corresponding y value, completing the critical point as (1/10, 3/10, 1/2).

Determining whether this critical point represents a maximum value can involve further steps such as using the second derivative test or considering the nature of the problem. For our function, this point yielded the maximum value f(1/10, 3/10, 1/2) = 9/20, solving the optimization problem efficiently.