Math Problem Solver

### **Solution to Find Local Maxima, Minima, and Saddle Points of $f(x, y) = 2x^2 - 8xy + y^4 - 4y^3$**

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#### **Step 1: Compute the First Partial Derivatives**

To find critical points, we first compute the partial derivatives of $f(x, y)$ with respect to $x$ and $y$:

1. **Partial derivative w.r.t. $x$:**

$$f_x = \frac{\partial f}{\partial x} = 4x - 8y$$

2. **Partial derivative w.r.t. $y$:**

$$f_y = \frac{\partial f}{\partial y} = -8x + 4y^3 - 12y^2$$

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#### **Step 2: Find Critical Points**

Critical points occur where $f_x = 0$ and $f_y = 0$.

1. **From $f_x = 0$:**

$$4x - 8y = 0 \implies x = 2y$$

2. **Substitute $x = 2y$ into $f_y = 0$:**

$$-8(2y) + 4y^3 - 12y^2 = 0 \implies -16y + 4y^3 - 12y^2 = 0$$

Simplify by dividing by 4:

$$-4y + y^3 - 3y^2 = 0 \implies y(y^2 - 3y - 4) = 0$$

Factor:

$$y(y - 4)(y + 1) = 0$$

So, the solutions are:

$$y = 0, \quad y = 4, \quad y = -1$$

Corresponding $x$-values:

- If $y = 0$, $x = 2(0) = 0$: **Critical point $(0, 0)$**.

- If $y = 4$, $x = 2(4) = 8$: **Critical point $(8, 4)$**.

- If $y = -1$, $x = 2(-1) = -2$: **Critical point $(-2, -1)$**.

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#### **Step 3: Compute the Second Partial Derivatives**

To classify the critical points, we compute the second partial derivatives:

1. $f_{xx} = \frac{\partial^2 f}{\partial x^2} = 4$

2. $f_{yy} = \frac{\partial^2 f}{\partial y^2} = 12y^2 - 24y$

3. $f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -8$

The **Hessian determinant** $D$ is:

$$D = f_{xx} f_{yy} - (f_{xy})^2 = 4(12y^2 - 24y) - (-8)^2 = 48y^2 - 96y - 64$$

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#### **Step 4: Classify Each Critical Point**

1. **Critical Point $(0, 0)$:**

- Compute $D$ at $y = 0$:

$$D = 48(0)^2 - 96(0) - 64 = -64$$

- Since $D < 0$, $(0, 0)$ is a **saddle point**.

2. **Critical Point $(8, 4)$:**

- Compute $D$ at $y = 4$:

$$D = 48(16) - 96(4) - 64 = 768 - 384 - 64 = 320$$

- Since $D > 0$ and $f_{xx} = 4 > 0$, $(8, 4)$ is a **local minimum**.

- Compute $f(8, 4)$:

$$f(8, 4) = 2(64) - 8(8)(4) + 256 - 256 = 128 - 256 + 0 = -128$$

3. **Critical Point $(-2, -1)$:**

- Compute $D$ at $y = -1$:

$$D = 48(1) - 96(-1) - 64 = 48 + 96 - 64 = 80$$

- Since $D > 0$ and $f_{xx} = 4 > 0$, $(-2, -1)$ is a **local minimum**.

- Compute $f(-2, -1)$:

$$f(-2, -1) = 2(4) - 8(-2)(-1) + 1 - 4(-1) = 8 - 16 + 1 + 4 = -3$$

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#### **Step 5: Summary of Results**

- **Local Minima:**

- At $(8, 4)$, $f(8, 4) = -128$.

- At $(-2, -1)$, $f(-2, -1) = -3$.

- **Saddle Point:**

- At $(0, 0)$.

- **Local Maxima:**

- None (the function has no local maxima in the observed domain).

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### **Final Answer**

- **Local minima:** $(8, 4)$ and $(-2, -1)$.

- **Saddle point:** $(0, 0)$.

- **No local maxima** exist for this function.

**Visualization Note:**

To confirm these findings, it is helpful to graph the function around these critical points using a 3D plotting tool (e.g., Matplotlib, GeoGebra). This will visually verify the presence of the two local minima and the saddle point.