Question

In Exercises 51 and \(52,\) find the critical points of the function and, from the form of the function, determine whether a relative maximum or a relative minimum occurs at each point. $$ f(x, y, z)=x^{2}+(y-3)^{2}+(z+1)^{2} $$

Short answer

The function has a critical point at (0, 3, -1) and it is a relative minimum.

Step-by-step solution

Compute the partial derivatives

The first step is to compute the partial derivatives of the function with respect to \(x\), \(y\), and \(z\). They are given by: \[ \frac{df}{dx} = 2x, \\ \frac{df}{dy} = 2(y - 3), \\ \frac{df}{dz} = 2(z + 1). \]

Find the critical points

A critical point occurs where all the partial derivatives of the function are zero. Setting the above derivatives to zero gives the equations: \[ 2x = 0, \\ 2(y - 3) = 0, \\ 2(z + 1) = 0. \] Solving these equations for \(x\), \(y\), and \(z\) yields: \(x = 0\), \(y = 3\), and \(z = -1\). Thus, the function has a critical point at (0, 3, -1).

Determine if the critical point is a relative maximum or a minimum

The form of \(f(x, y, z) = x^{2}+(y-3)^{2}+(z+1)^{2}\) resembles the equation of a sphere shifted from the origin, which is always non-negative (i.e., \( \geq 0\)). The minimum value of this function is 0, which happens at the point (0, 3, -1). So, this critical point is a relative minimum point of the function.

Key concepts

Partial Derivatives

Understanding partial derivatives is a foundational aspect of multivariable calculus. These derivatives form the basis for identifying the trends and behaviors of functions that depend on more than one variable. Unlike in single-variable calculus where the derivative gives the rate of change of a function in relation to one variable, partial derivatives measure how a function changes as each variable is varied separately while the others are held constant.

For example, the exercise provided involves a function of three variables, f(x, y, z). The partial derivative of this function with respect to x is found by treating y and z as constants and differentiating with respect to x, leading to \( \frac{df}{dx} = 2x \). Similarly, the derivatives with respect to y and z are \( \frac{df}{dy} = 2(y - 3) \) and \( \frac{df}{dz} = 2(z + 1) \) respectively. These derivatives are used to determine the rates of change in the function in the direction of each individual variable.

Relative Maximum and Minimum

The concept of relative maximum and minimum in multivariable calculus extends the idea of finding peaks and troughs in a graph from single-variable calculus to functions of several variables. A relative maximum is a point at which a function reaches a local peak, whereas a relative minimum is where it dips to a local low point, compared to the surrounding points.

The given function f(x, y, z) has a critical point where the partial derivatives all equate to zero. This is a potential location for a relative maximum or minimum. The nature of the function's graph around this point helps in categorizing it. In the exercise, the squared terms suggest that the function has a bowl-shaped graph, indicating the critical point is likely a relative minimum since the function takes on its smallest value at this point.

Multivariable Calculus

The branch of mathematics dealing with functions that rely on several variables is known as multivariable calculus. It involves generalizing the principles of calculus to higher dimensions. Tasks like finding maxima and minima, sketching the shapes of functions, and predicting the behavior of systems that change across more than one dimension are all within its purview.

In multivariable calculus, a critical point of a function of several variables is where the gradient of the function is zero or undefined. The gradient is a vector composed of all the partial derivatives with respect to each independent variable. Identifying critical points, as in the provided exercise, can reveal a lot about the function's geometry, such as the location of hills, valleys, and saddle points on the graph of the function. The critical point at \( (0, 3, -1) \) in this exercise is found to be a relative minimum by analyzing the positive squared terms in the function, signifying that it's the lowest point in its immediate vicinity.