Problem 46 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of the following functions on the given set \(R\). $$\begin{aligned} &f(x, y)=6-x^{2}-4 y^{2}\\\ &R=\\{(x, y):-2 \leq x \leq 2,-1 \leq y \leq 1\\} \end{aligned}$$

Step-by-Step Solution

Answer

Answer: The absolute maximum value is 6, and the absolute minimum value is 0.

Step 1: Compute the partial derivatives

To find the critical points of the function, we will compute the first partial derivatives with respect to both variables x and y: $$\begin{aligned} \frac{\partial f}{\partial x} &= -2x \\ \frac{\partial f}{\partial y} &= -8y \end{aligned}$$

Step 2: Calculate critical points in the interior

Now, set the derivatives equal to zero and solve for the critical points: $$\begin{aligned} -2x &= 0 \implies x = 0 \\ -8y &= 0 \implies y = 0 \end{aligned}$$ Thus, we have a single critical point in the interior at \((0, 0)\). Now, we will check if it falls within our region R.

Step 3: Verify if the critical point is within the region R

Recall that our region R is \(\{(x, y) : -2 \leq x \leq 2, -1 \leq y \leq 1\}\). The critical point we found at \((0, 0)\) falls within this region, so we will consider it when finding the absolute extrema.

Step 4: Analyze the function on the boundary of R

The boundary of R is composed of 4 lines: \(x=-2, x=2, y=-1\), and \(y=1\). Analyze \(f(x, y)\) on each of these lines by substituting the corresponding values and simplifying the respective expressions: 1. On the line \(x=-2\): $$f(-2, y) = 6 - (-2)^2 - 4y^2 = 2 - 4y^2$$ 2. On the line \(x=2\): $$f(2, y) = 6 - (2)^2 - 4y^2 = 2 - 4y^2$$ 3. On the line \(y=-1\): $$f(x, -1) = 6 - x^2 - 4(-1)^2 = 2 - x^2$$ 4. On the line \(y=1\): $$f(x, 1) = 6 - x^2 - 4(1)^2 = 2 - x^2$$

Step 5: Calculate function values at corners and critical point

The corners of region R are \((-2, -1), (-2, 1), (2, -1),\) and \((2, 1)\). Calculate the value of the function \(f(x, y)\) at these corners and the critical point (0,0): 1. \(f(-2, -1) = 2 - 4(-1)^2 = 0\) 2. \(f(-2, 1) = 2 - 4(1)^2 = 0\) 3. \(f(2, -1) = 2 - 4(-1)^2 = 0\) 4. \(f(2, 1) = 2 - 4(1)^2 = 0\) 5. \(f(0, 0) = 6 - 0^2 - 4(0)^2 = 6\)

Step 6: Determine the absolute maximum and minimum values

Comparing the function values at corners and the critical point, we find that the absolute maximum is \(f(0, 0) = 6\), and the absolute minimum is \(f(-2, -1) = f(-2, 1) = f(2, -1) = f(2, 1) = 0\).

Key Concepts

Partial Derivatives

Partial derivatives are a key concept in multivariable calculus. They represent the rate of change of a function with respect to one of its variables while keeping the other variables constant.

For a function of two variables like \(f(x, y)\), the partial derivative with respect to \(x\) is denoted as \(\frac{\partial f}{\partial x}\), and with respect to \(y\) as \(\frac{\partial f}{\partial y}\).

To find partial derivatives, differentiate the function treating all other variables as constants:

• For \(\frac{\partial f}{\partial x} = -2x\), the function is differentiated with respect to \(x\) keeping \(y\) constant.
• For \(\frac{\partial f}{\partial y} = -8y\), the function is differentiated with respect to \(y\) keeping \(x\) constant.

These derivatives are used to find critical points, important when optimizing functions.

Critical Points

Critical points are where the function's partial derivatives equal zero or are undefined.

This is crucial for finding potential maximum or minimum values of the function.
Set the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) to zero:

• \(-2x = 0\) yields \(x = 0\)
• \(-8y = 0\) yields \(y = 0\).

Thus, we get the critical point \((0, 0)\).

Always check if it lies within the given set \(R\), as critical points outside \(R\) are not relevant for finding the extrema within \(R\).

Absolute Extrema

Absolute extrema refer to the highest (maximum) and lowest (minimum) values of a function over a given region.

To find them, you must evaluate the function at critical points and along the boundaries of the region.

In our exercise, we discovered that the function value at the critical point

• \(f(0, 0) = 6\) represents the absolute maximum.
• While function values at any corner of the region \((2, 1), (2, -1), (-2, 1), (-2, -1)\) are all 0, which is the absolute minimum.

This means comparing values at critical points and boundaries gives you a clear picture of the function's range within a region.

Boundary Analysis

Boundary analysis involves checking the behavior of the function at the edges of the region it is confined to.

For the rectangle \(R\), the edges or boundaries consist of specific lines where \(x\) or \(y\) is constant.

Evaluate the function along these lines:

• \(x = -2, f(-2, y) = 2 - 4y^2\)
• \(x = 2, f(2, y) = 2 - 4y^2\)
• \(y = -1, f(x, -1) = 2 - x^2\)
• \(y = 1, f(x, 1) = 2 - x^2\)

Then calculate \(f(x, y)\) at the corners: \((-2, -1), (-2, 1), (2, -1), (2, 1)\).

Boundary evaluation ensures you capture all potential extrema outside of just the interior critical points.