Problem 45 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of the following functions on the given set \(R\). $$\begin{aligned} &f(x, y)=4+2 x^{2}+y^{2}\\\ &R=\\{(x, y):-1 \leq x \leq 1,-1 \leq y \leq 1\\} \end{aligned}$$

Step-by-Step Solution

Answer

Answer: The absolute maximum value of the function is 7, and the absolute minimum value is 4.

Step 1: Find Partial Derivatives

First, find the partial derivative of the function with respect to \(x\) and \(y\). We'll need these to identify the critical points: $$\frac{\partial f}{\partial x} = 4x \quad \text{and} \quad \frac{\partial f}{\partial y} = 2y$$ Now, set both partial derivatives equal to zero to find the potential critical points: $$4x=0 \quad \text{and} \quad 2y=0$$ This leads to the critical point of \((x, y)=(0, 0)\).

Step 2: Evaluate Function at Critical Points and Boundary Points

Now, we need to evaluate the function at the critical point \((0,0)\), as well as the boundary points. We can use the equation \(f(x,y)=4+2x^2+y^2\) for this: $$f(0,0)=4+2(0)^2+(0)^2=4$$ Next, evaluate the function at the boundary points. Since the region is a square with vertices at \((-1,-1)\), \((-1,1)\), \((1,-1)\), and \((1,1)\), evaluate the function at these points: $$f(-1,-1)=4+2(-1)^2+(-1)^2=7$$ $$f(-1,1)=4+2(-1)^2+(1)^2=7$$ $$f(1,-1)=4+2(1)^2+(-1)^2=7$$ $$f(1,1)=4+2(1)^2+(1)^2=7$$

Step 3: Determine Absolute Maximum and Minimum Values

Now we have all the function values at the critical point and boundary points, we can identify the absolute maximum and minimum values: - At the critical point \((0, 0)\), \(f(x, y) = 4\) - On the boundary points, \(f(x, y) = 7\) Since both of these values are unique, we can conclude: - The absolute minimum value of the function is \(f(0,0)=4\) - The absolute maximum value of the function is \(f(-1,-1)=f(-1,1)=f(1,-1)=f(1,1)=7\)

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function changes as its variables change independently. In this exercise, we have a function of two variables: \(f(x, y)=4+2x^2+y^2\). To explore how this function behaves, we need its partial derivatives with respect to \(x\) and \(y\).

• The partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = 4x\).
• The partial derivative with respect to \(y\) is \(\frac{\partial f}{\partial y} = 2y\).

To find these derivatives, we treat all other variables except the one we are differentiating against as constants. Partial derivatives are crucial for locating critical points, where the function might reach a maximum or minimum within a given domain. To identify such points, we set these derivatives equal to zero and solve for \(x\) and \(y\). In this case, we get the potential critical point at \((0,0)\).

Critical Points

Critical points of a function are where its partial derivatives are zero or undefined. They are candidates for local minima, maxima, or saddle points of the function.
In our function \(f(x, y)=4+2x^2+y^2\), we set the partial derivatives to zero:

• \(4x = 0\) → \(x = 0\)
• \(2y = 0\) → \(y = 0\)

The resulting critical point is \((0, 0)\).We evaluate the function at this critical point,
\(f(0,0) = 4\), to determine its behavior there. However, critical points alone do not indicate whether we have a maximum or minimum; thus, we must also consider the function's behavior on the boundaries of the region \(R\).

Boundary Evaluation

Evaluating a function on the boundary of its domain ensures that no potential extremums are missed. In this exercise, the set \(R\) is defined as a square \(-1 \leq x \leq 1\) and \(-1 \leq y \leq 1\). To find values on the boundary:We check the vertices of the square:

• \((-1, -1)\): \(f(-1, -1) = 7\)
• \((-1, 1)\): \(f(-1, 1) = 7\)
• \((1, -1)\): \(f(1, -1) = 7\)
• \((1, 1)\): \(f(1, 1) = 7\)

By evaluating these boundary points, we ensure that the absolute extrema are indeed identified within and on the edges of the region \(R\). The boundary points reveal a consistent function value of 7, indicating potential maxima. This process confirms that all relevant points are considered when determining the absolute extremes of the function.

Absolute Maximum and Minimum

After identifying critical points and evaluating the function on its boundaries, we determine the absolute maximum and minimum values on the given set \(R\). An absolute maximum is the highest value across the entire function domain, and an absolute minimum is the lowest.In our case, the function value:

• At the critical point \((0, 0)\), is \(f(0,0) = 4\), suggesting it as a minimum.
• Across all evaluated boundary points, \(f(x, y) = 7\), suggesting these are maxima.

Thus, the absolute minimum of the function is \(4\), and the absolute maximum is \(7\). These results demonstrate how critical points and boundary evaluations complement each other in pinpointing the extrema within a defined region.