Questions: Find the absolute maximum and minimum values of the following function on the given set R. f(x, y) = x^2 + y^2 - 2y + 1 ; R = (x, y): x^2 + y^2 ≤ 9 What is the absolute maximum value? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The absolute maximum value is . (Simplify your answer.) B. There is no absolute maximum value.

Find the absolute maximum and minimum values of the following function on the given set R.
f(x, y) = x^2 + y^2 - 2y + 1 ; R = (x, y): x^2 + y^2 ≤ 9

What is the absolute maximum value? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The absolute maximum value is . (Simplify your answer.)
B. There is no absolute maximum value.

Transcript text: Find the absolute maximum and minimum values of the following function on the given set $R$. \[ f(x, y)=x^{2}+y^{2}-2 y+1 ; R=\left\{(x, y): x^{2}+y^{2} \leq 9\right\} \] What is the absolute maximum value? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The absolute maximum value is $\square$ . (Simplify your answer.) B. There is no absolute maximum value.

Solution

Solution Steps

To find the absolute maximum and minimum values of the function $ f(x, y) = x^2 + y^2 - 2y + 1 $ on the set $ R = \{(x, y) : x^2 + y^2 \leq 9\} $, we need to:

Evaluate the function at the critical points inside the region $ R $.
Evaluate the function on the boundary of the region $ R $, which is the circle $ x^2 + y^2 = 9 $.
Compare these values to determine the absolute maximum and minimum.

Step 1: Identify the Function and the Region

We are given the function: \[ f(x, y) = x^2 + y^2 - 2y + 1 \] and the region: \[ R = \{(x, y) : x^2 + y^2 \leq 9\} \] This region is a disk of radius 3 centered at the origin.

Step 2: Find Critical Points Inside the Region

To find the critical points, we need to find where the gradient of $ f $ is zero. The gradient is given by: \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] Calculating the partial derivatives: \[ \frac{\partial f}{\partial x} = 2x \] \[ \frac{\partial f}{\partial y} = 2y - 2 \] Setting these equal to zero to find the critical points: \[ 2x = 0 \implies x = 0 \] \[ 2y - 2 = 0 \implies y = 1 \] Thus, the critical point is $(0, 1)$.

Step 3: Evaluate the Function at the Critical Point

Substitute $(0, 1)$ into the function: \[ f(0, 1) = 0^2 + 1^2 - 2(1) + 1 = 1 - 2 + 1 = 0 \]

Step 4: Evaluate the Function on the Boundary

The boundary of the region is given by $ x^2 + y^2 = 9 $. We can parameterize this boundary using: \[ x = 3 \cos \theta, \quad y = 3 \sin \theta \] Substitute these into the function: \[ f(3 \cos \theta, 3 \sin \theta) = (3 \cos \theta)^2 + (3 \sin \theta)^2 - 2(3 \sin \theta) + 1 \] \[ = 9 \cos^2 \theta + 9 \sin^2 \theta - 6 \sin \theta + 1 \] Using the Pythagorean identity $\cos^2 \theta + \sin^2 \theta = 1$: \[ = 9(1) - 6 \sin \theta + 1 = 10 - 6 \sin \theta \] The maximum and minimum values of $ -6 \sin \theta $ occur at $\sin \theta = 1$ and $\sin \theta = -1$: \[ \text{Maximum: } 10 - 6(-1) = 10 + 6 = 16 \] \[ \text{Minimum: } 10 - 6(1) = 10 - 6 = 4 \]