Question

In Exercises \(53-62,\) find the absolute extrema of the function over the region \(R .\) (In each case, \(R\) contains the boundaries.) Use a computer algebra system to confirm your results. $$ \begin{aligned} &f(x, y)=12-3 x-2 y\\\ &R: \text { The triangular region in the } x y \text { -plane with vertices }(2,0),\\\ &(0,1), \text { and }(1,2) \end{aligned} $$

Short answer

The absolute maximum of the function \(f(x, y)=12-3x-2y\) over the region \(R\) is 10 at the point (0,1) and the absolute minimum is 6 at the points (2,0) and (1,2).

Step-by-step solution

Defining the function and the region

Firstly, the function and the region in question are defined. The function in question is \(f(x, y)=12-3x-2y\). The region \(R\) is a triangular region in the xy-plane, defined by the vertices (2,0), (0,1), and (1,2). The region includes these points and all points within the triangle formed by them.

Finding Critical Points

Critical points of function \(f\) are found by setting the partial derivatives of \(f\) to zero. The partial derivative of \(f\) with respect to \(x\) is -3 and with respect to \(y\) is -2. Since these are constant, there are no critical points for this function within the region.

Checking boundaries

Because there are no critical points, the absolute maximum and minimum must occur on the boundaries of the region. Evaluation of the function at the vertices of the triangle yields the function values: \(f(2,0)=12-3(2)-2(0)=6, f(0,1)=12-3(0)-2(1)=10, f(1,2) = 12-3(1)-2(2)=6.\) These are the only possible points for absolute maxima and minima.

Determine the absolute extrema

After evaluating the function at the vertices of the triangular region, it can be seen that the maximum function value is 10 at (0,1) and the minimum value is 6 at both the points (2,0) and (1,2). So, the absolute maximum is 10 and the absolute minimum is 6 in the given region.

Key concepts

Critical Points

When we discuss critical points in the context of multivariable calculus, we're looking at where the function's rate of change is temporarily halted—in other words, where it flattens out. To find these, we calculate the function's partial derivatives and look for points where these derivatives are both zero. This situation suggests a potential local maxima or minima, or saddle points.
For the given function, critical points would typically be found by setting the partial derivatives with respect to both variables, in this case, x and y, to zero. However, as the partial derivatives of the provided function, \( -3 \text{ and } -2 \) neither depend on x nor y, they are constant. This tells us something important: since neither derivative is ever zero, there are no interior critical points for this function within the triangular region.
This lack of critical points directs us to the boundaries of the region to search for absolute extrema. When giving this content to students, emphasis on understanding the nature of partial derivatives and their constancy is paramount, because it significantly affects how and where extrema can be found.

Partial Derivatives

Partial derivatives act as a compass that tells us the direction in which a function is increasing or decreasing most rapidly at any given point, but only along axes in x or y, not diagonally or in mixed directions. Think of them as slices of the function's rate of change in one dimension at a time.
In this function \( f(x, y)=12-3x-2y \), the partial derivative with respect to x is \( \frac{\partial f}{\partial x} = -3 \) and with respect to y is \( \frac{\partial f}{\partial y} = -2 \). This means that for every unit you move in the x-direction, the function decreases by three units, and for every unit you move in the y-direction, it decreases by two units, irrespective of your position in the xy-plane.
Understanding partial derivatives is crucial because they can lead us to critical points, which are instrumental in identifying local maxima and minima of functions. Here, however, these derivatives also reveal that our function has a consistent slope with no variation—thus no critical points emerge from differentiating.

Maxima and Minima

Maxima and minima of a function, also known as extrema, are the highest and lowest points respectively, where a function reaches its maximum and minimum values. When we have a bounded region like the triangular area in our exercise, we look for absolute extrema, meaning the highest and lowest values the function can possibly take within this specific zone.
With no critical points inside the triangle, we turn our attention towards the edges and vertices of the triangle. It's possible for absolute extrema to occur at these boundaries, so we evaluate the function at the vertices (the points provided) to find that \( f(2,0)=6 \) and \( f(1,2)=6 \) (minimum values) while \( f(0,1)=10 \) (maximum value).
In teaching this concept, it's important to highlight the significance of evaluating the function not just at critical points, but also along the boundaries of a region, especially when working with functions of several variables. By understanding where to look for extrema and how to evaluate them, students can effectively determine the solutions to a wide range of real-world optimization problems.