Question

If possible, find the absolute maximum and minimum values of the following functions on the set \(R\). $$f(x, y)=2 e^{-x-y} ; R=\\{(x, y): x \geq 0, y \geq 0\\}$$

Short answer

Answer: The absolute maximum value is \(2\) at the point \((0,0)\), and the absolute minimum value is \(0\) as either \(x \to \infty\) or \(y \to \infty\).

Step-by-step solution

Determine critical points

To find the critical points, we need to take partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\). Then, we'll look for points where both partial derivatives equal zero or are undefined. $$f_x = \frac{\partial f}{\partial x} = -2e^{-x-y}$$ $$f_y = \frac{\partial f}{\partial y} = -2e^{-x-y}$$ Since the partial derivatives are never undefined, we will set both \(f_x\) and \(f_y\) to zero and solve for \(x\) and \(y\): $$-2e^{-x-y} = 0 \Rightarrow e^{-x-y} = 0 $$ However, the exponential function is always greater than 0, so there is no solution in this case. This tells us that there are no critical points within the region \(R\).

Evaluate the function at the boundaries

Since we found no critical points, we must evaluate the function at the boundaries of our region \(R\), which is specified by \(x \geq 0, y \geq 0\). There are 3 boundaries to consider: 1. \(x = 0\) and \(y \geq 0\) 2. \(y = 0\) and \(x \geq 0\) 3. \(x \to \infty\) and/or \(y \to \infty\) For boundary (1), the function becomes: $$f(0, y) = 2e^{-y}$$ Since \(y \geq 0\), the minimum value occurs at \(y = 0\), and the maximum value occurs as \(y \to \infty\). For boundary (2), the function becomes: $$f(x, 0) = 2e^{-x}$$ Similarly, the minimum value occurs at \(x = 0\) and the maximum value occurs as \(x \to \infty\). For boundary (3), the function becomes: $$f(x, y) = 2e^{-x-y}$$ As \(x \to \infty\), or \(y \to \infty\), the exponential function approaches 0. Thus, the minimum value occurs at \(x \to \infty\) and/or \(y \to \infty\).

Find absolute maximum and minimum values

After evaluating the function at the boundaries, we can determine the absolute maximum and minimum values. The maximum value of the function occurs at \((0,0)\): $$f(0, 0) = 2e^{0} = 2$$ The minimum value occurs at the boundary where either \(x \to \infty\) or \(y \to \infty\): $$\lim_{x \to \infty, y \to \infty} 2e^{-x-y} = 0$$ Thus, the absolute maximum value is \(2\) at the point \((0,0)\), and the absolute minimum value is \(0\) as either \(x \to \infty\) or \(y \to \infty\).

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, providing information about how a function changes as its variables change individually. In our specific function, \(f(x, y) = 2 e^{-x-y}\), partial derivatives allow us to explore how the function behaves as either \(x\) or \(y\) varies, while the other variable is held constant.

To find the partial derivative with respect to \(x\), denoted as \(f_x\), we differentiate \(f\) with respect to \(x\) while treating \(y\) as a constant. Similarly, finding the partial derivative with respect to \(y\) involves treating \(x\) as constant. This yields:
\[ f_x = \frac{\partial f}{\partial x} = -2e^{-x-y} \]
\[ f_y = \frac{\partial f}{\partial y} = -2e^{-x-y} \]
Neither of these partial derivatives ever equals zero for any real numbers \(x\) or \(y\), indicating that the function decreases as either variable increases. This results from the constant negative multiplier \(-2\) in the partial derivatives, confirming a consistent rate of decay for the exponentials.

Critical Points

Critical points are locations where the function's rate of change is either zero or undefined, making them potential candidates for local extrema. In our case, finding critical points means solving where both partial derivatives of \(f(x, y) = 2 e^{-x-y}\) equal zero.

We set \(f_x = 0\) and \(f_y = 0\). However, because each partial derivative simplifies to \(-2e^{-x-y}\), they cannot equal zero since exponentials are positive for all real inputs. This result signifies there are no critical points inside the region \(R = \{(x, y): x \geq 0, y \geq 0\}\).

Understanding critical points helps us discern function behavior within defined domains, but sometimes, like for exponential functions, boundaries need to be evaluated to find extrema.

Boundary Evaluation

Boundary evaluation is vital when critical points are elusive, as it helps determine the behavior of a function along the edges of the region. In this scenario, the region is confined to the first quadrant, \(x \geq 0\) and \(y \geq 0\). Thus, we assess how the function behaves along these boundaries:

• For \(x = 0\): The function simplifies to \(f(0, y) = 2e^{-y}\). As \(y\) increases, \(f(0, y)\) decreases, maxing at \(y = 0\) and minning as \(y \to \infty\).
• For \(y = 0\): The function is \(f(x, 0) = 2e^{-x}\), showing similar behavior where it decreases as \(x\) increases, with a max at \(x = 0\).
• When both \(x\) and \(y\) grow to infinity: The function tends to zero, since \(f(x, y) = 2e^{-x-y}\) implies the exponential decays rapidly with increasing \(x\) and \(y\).

Boundary evaluation is thus crucial for ensuring all possible extrema are checked. Here, the absolute maximum of 2 is found at \((0,0)\), and the minimum approaches zero as one or both variables become infinitely large.