Question

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. $$ f(x, y)=x^{2}+x y, \quad R=\\{(x, y):|x| \leq 2,|y| \leq 1\\} $$

Short answer

The function \(f(x, y) = x^{2} + xy\) achieves maximum value of 6 at points (-2,1) and (2,1), and minimum value of 2 at points (-2,-1) and (2,-1), over the given region \(R\), where \(|x| \leq 2, |y| \leq 1\).

Step-by-step solution

Find the critical points

Critical points occur where the first derivative equals zero, or is undefined. However, because \(f(x, y)\) is a polynomial, there are no points where the derivative is undefined. So, calculate the first derivatives of the function with respect to both \(x\) and \(y\): \[f_{x}(x,y) =2x+y\] \[f_{y}(x,y) =x\] Set the first derivatives equal to zero and solve the system of equations: \[2x+y=0\] \[x=0\] solving this, we get (0,0).

Evaluate \(f(x, y)\) at the critical points

The next step is to evaluate the function at the critical point (0,0). Substituting (0,0) into \(f(x, y)\), we get \(f(0,0) = 0\).

Evaluate \(f(x, y)\) on the boundary of \(R\)

The boundary of the region \(R\) where \(|x|\leq2\), \(|y|\leq1\) corresponds to four line segments: \(x=\pm 2, -1\leq y\leq1\) and \(y=\pm 1, -2\leq x\leq2\). For \(x=2\), \(f(2,y) = 4+2y\), and for \(x=-2\), \(f(-2,y) = 4-2y\). For \(y=1\), \(f(x,1) = x^{2}+x\), and for \(y=-1\), \(f(x,-1) = x^{2}-x\). By analysing these four functions, we see that the absolute extrema must occur at \(x = \pm2\) for each \(y\) value, and at \(y = \pm1\) at each \(x\) value. Therefore, examine the endpoints: \[f(2,1) = 6, f(-2,1) = 6, f(2,-1) = 2, f(-2,-1) = 2, f(-2,1) = 6, f(-2,-1) = 2\].

Identify the maximum and minimum values

Now, compare all the function values obtained in Steps 2 and 3 to determine the maximum and minimum. Having calculated, the maximum occurs at points (-2,1), (2,1) where \(f(x,y) = 6\). The minimum occurs at points (-2,-1), (2,-1) where \(f(x,y) = 2\).

Key concepts

Critical Points in Calculus

When exploring functions in calculus, critical points hold key information regarding the function's behaviour. In the context of finding absolute extrema, critical points are crucial as they could potentially be points where the function reaches its highest or lowest value.

Mathematically, critical points are locations on the graph of a function where the first derivative is either zero or undefined, indicating a potential local maximum, local minimum, or inflection point. In multivariable calculus, like with the function f(x, y) = x^2 + xy from the exercise, the process involves taking partial derivatives of the function with respect to each variable and setting these derivatives equal to zero. Solving the resulting system of equations will provide the critical points in the domain of the function.

First Derivatives

The first derivative of a function is the backbone of the calculus concept known as differentiation. It tells us about the rate at which the function's value is changing at any given point - essentially, it's the slope of the function at that point.

In multivariable functions, like our function f(x, y), we use partial derivatives to learn about changes in the function with respect to each variable separately. These first partial derivatives can be used to locate critical points by finding where they are simultaneously zero. Keep in mind, the function’s extrema could be hidden among these critical points, and finding them is an essential step in optimization.

Boundary Evaluation

Why Evaluate Boundaries?

Boundary evaluation is as pivotal as finding critical points when looking for absolute extrema. Since critical points only provide information about the interior of the region, evaluating the function's boundary ensures that no extreme values are disregarded. This becomes particularly pertinent when the domain of the function is restricted, as is the case with the function f(x, y) in the given region R.

Consider the edges of the given domain and calculate the function values at these edges to exhaustively search for maximum and minimum values. By doing so, we make sure that every possible candidate for absolute extrema within the given region R is accounted for, ensuring a reliable result for our optimization endeavours.

Optimization in Multivariable Calculus

Finding Extrema in Multiple Dimensions

Optimization in multivariable calculus may sound daunting, but it’s a natural extension of single-variable optimization into higher dimensions. Here, not only do we examine peaks and valleys (maxima and minima), we also navigate across surfaces and contours defined by functions with more than one variable.

In optimization, the goal is to find the best (optimal) values according to a given criteria – in our case, the highest and lowest points on a surface defined by f(x, y). The process involves finding critical points, evaluating the function at those points, investigating the boundaries, and finally considering all this information collectively to pinpoint where the function achieves its absolute extremum within the specified region. Multivariable functions can yield multiple extrema which are vital in many sectors like engineering and economics for maximizing efficiency or minimizing cost.