Find the local extrema of

f(x,y)=2⋅y2−16⋅y+2⋅x2−15⋅x+1

[x0,y0]=

[15/4,4]

Below is your answer, as understood by the system:

[154,4]

Determine whether it is a local minimum point, a local maximum point, a saddle point or it cannot be determined using the second derivative test:


Minimum

Determine the largest and smallest value that the function accepts in the domain

x2+y2≤25

Min=


max=

### 1. **Finding Local Extrema Using Second Derivative Test**

Given the function:

\[
f(x, y) = 2y^2 - 16y + 2x^2 - 15x + 1
\]

We need to determine the nature of the point \(\left(\frac{15}{4}, 4\right)\).

#### Step 1: Calculate the First Partial Derivatives

The first partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\) are:

\[
f_x = \frac{\partial f}{\partial x} = 4x - 15
\]
\[
f_y = \frac{\partial f}{\partial y} = 4y - 16
\]

Setting these equal to zero to find the critical points:

\[
4x - 15 = 0 \quad \Rightarrow \quad x = \frac{15}{4}
\]
\[
4y - 16 = 0 \quad \Rightarrow \quad y = 4
\]

Thus, \(\left(\frac{15}{4}, 4\right)\) is a critical point.

#### Step 2: Calculate the Second Partial Derivatives

The second partial derivatives are:

\[
f_{xx} = \frac{\partial^2 f}{\partial x^2} = 4
\]
\[
f_{yy} = \frac{\partial^2 f}{\partial y^2} = 4
\]
\[
f_{xy} = f_{yx} = 0
\]

#### Step 3: Apply the Second Derivative Test

The discriminant \(D\) is given by:

\[
D = f_{xx} \cdot f_{yy} - (f_{xy})^2 = 4 \cdot 4 - 0^2 = 16
\]

Since \(D > 0\) and \(f_{xx} > 0\), the point \(\left(\frac{15}{4}, 4\right)\) is a **local minimum**.

### 2. **Determining the Largest and Smallest Values in the Domain \(x^2 + y^2 \leq 25\)**

The domain is the disk defined by \(x^2 + y^2 \leq 25\).

#### Step 1: Evaluate \(f(x, y)\) at the Boundary \(x^2 + y^2 = 25\)

Substitute \(x = 5\cos\theta\) and \(y = 5\sin\theta\):

\[
f(5\cos\theta, 5\sin\theta) = 2(5\sin\theta)^2 - 16(5\sin\theta) + 2(5\cos\theta)^2 - 15(5\cos\theta) + 1
\]

Simplifying:

\[
f(5\cos\theta, 5\sin\theta) = 50(\sin^2\theta + \cos^2\theta) - 80\sin\theta - 75\cos\theta + 1
\]

Since \(\sin^2\theta + \cos^2\theta = 1\):

\[
f(5\cos\theta, 5\sin\theta) = 50 - 80\sin\theta - 75\cos\theta + 1
\]
\[
f(5\cos\theta, 5\sin\theta) = 51 - 80\sin\theta - 75\cos\theta
\]

The extrema on the boundary can be found by evaluating the above expression for different values of \(\theta\).

#### Step 2: Evaluate \(f(x, y)\) at the Local Minimum Point

At the local minimum point \(\left(\frac{15}{4}, 4\right)\):

\[
f\left(\frac{15}{4}, 4\right) = 2(4)^2 - 16(4) + 2\left(\frac{15}{4}\right)^2 - 15\left(\frac{15}{4}\right) + 1
\]

Simplifying:

\[
f\left(\frac{15}{4}, 4\right) = 32 - 64 + \frac{450}{8} - \frac{225}{4} + 1
\]
\[
f\left(\frac{15}{4}, 4\right) = -32 + \frac{450}{8} - \frac{450}{8} + 1 = 1
\]

Thus, the local minimum value within the domain is **1**.

#### Final Answer:

- **Minimum Value:** \(1\)
- **Maximum Value:** \(51 + 80 + 75 = 206\) (calculated from the boundary analysis at \(\sin\theta = -1\), \(\cos\theta = -1\))

Would you like more details on any of these calculations, or do you have any specific questions?

### Relative Questions:

1. How would you find local maxima using the second derivative test?
2. Why does \(D > 0\) indicate a minimum or maximum?
3. How would the result change if \(f_{xx} < 0\)?
4. What are the implications if \(D < 0\)?
5. Can a local minimum be also a global minimum?
6. How does boundary analysis work in multivariable optimization?
7. What if the constraint were \(x^2 + y^2 < 25\) instead of \(\leq\)?
8. How would Lagrange multipliers apply to this problem?

### Tip:

When analyzing functions within constrained domains, always check boundary conditions to ensure global extrema are correctly identified.

Learn how to find local extrema of the function f(x,y)=2y^2−16y+2x^2−15x+1 using the second derivative test. Determine whether the critical point (15/4, 4) is a local minimum, local maximum, saddle point, or undetermined. Understand the implications of the discriminant for the second derivative test. Explore boundary analysis to determine the largest and smallest values of the function within the domain x^2+y^2≤25.

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