Let's solve **Question 1** step-by-step: --- ## 1. (Section 10.2) Given the function: \[ f(x, y) = \frac{y}{x} - \ln x + \frac{1}{2}y^2 \] We are to **find the critical points** and **their nature**. ### **Step 1: Find the partial derivatives** #### \(\frac{\partial f}{\partial x}\): \[ f(x, y) = \frac{y}{x} - \ln x + \frac{1}{2}y^2 \] \[ \frac{\partial f}{\partial x} = -\frac{y}{x^2} - \frac{1}{x} \] #### \(\frac{\partial f}{\partial y}\): \[ \frac{\partial f}{\partial y} = \frac{1}{x} + y \] --- ### **Step 2: Set the partial derivatives equal to zero** \[ \frac{\partial f}{\partial x} = 0 \implies -\frac{y}{x^2} - \frac{1}{x} = 0 \] \[ \frac{\partial f}{\partial y} = 0 \implies \frac{1}{x} + y = 0 \] From the second equation: \[ y = -\frac{1}{x} \] Plug this value into the first equation: \[ -\frac{-\frac{1}{x}}{x^2} - \frac{1}{x} = 0 \] \[ \frac{1}{x^3} - \frac{1}{x} = 0 \] \[ \frac{1 - x^2}{x^3} = 0 \] So, \[ 1 - x^2 = 0 \implies x^2 = 1 \implies x = 1 \text{ or } x = -1 \] Now, plug back to find \(y\): - If \(x = 1\): \(y = -1\) - If \(x = -1\): \(y = 1\) So, the **critical points are**: \((1, -1)\) and \((-1, 1)\). --- ### **Step 3: Determine the nature of the critical points** Compute the **second partial derivatives**: \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{2y}{x^3} + \frac{1}{x^2} \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 1 \] \[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -\frac{1}{x^2} \] Now, calculate the **Hessian determinant** at each critical point: \[ H = f_{xx} f_{yy} - (f_{xy})^2 \] #### At \((1, -1)\): - \(x = 1, y = -1\) - \(f_{xx} = \frac{2(-1)}{1^3} + \frac{1}{1^2} = -2 + 1 = -1\) - \(f_{yy} = 1\) - \(f_{xy} = -1\) \[ H = (-1)(1) - (-1)^2 = -1 - 1 = -2 < 0 \] #### At \((-1, 1)\): - \(x = -1, y = 1\) - \(f_{xx} = \frac{2(1)}{(-1)^3} + \frac{1}{(-1)^2} = \frac{2}{-1} + 1 = -2 + 1 = -1\) - \(f_{yy} = 1\) - \(f_{xy} = -\frac{1}{(-1)^2} = -1\) \[ H = (-1)(1) - (-1)^2 = -1 - 1 = -2 < 0 \] --- ### **Conclusion:** - **Critical points:** \((1, -1)\) and \((-1, 1)\). - **Nature:** Since the Hessian determinant is negative at both points, both are **saddle points**. --- ## **Final Answer:** **The function \( f(x, y) = \frac{y}{x} - \ln x + \frac{1}{2}y^2 \) has two critical points: \((1, -1)\) and \((-1, 1)\). Both are saddle points.** --- Would you like to proceed with the next question?

Let's proceed with **Question 2**, based on the provided topics from sections 10.3, etc. Since the exact question isn't fully visible, I'll assume it's related to the same topics involving double integrals, set regions, or transformations. --- ## 2. (Section 10.3) Suppose the problem involves evaluating a double integral over a certain region or transforming variables. ### **Sample Problem:** **Evaluate the double integral**: \[ I = \iint_{R} xy \, dA \] where \( R \) is the region in the plane bounded by: \[ x \geq 0, \quad y \geq 0, \quad y = x^2, \quad y = 4x \] --- ### **Step 1: Describe the region \( R \)** - \( y = x^2 \) (parabola) - \( y = 4x \) (line) Find their intersection points: \[ x^2 = 4x \implies x^2 - 4x = 0 \implies x(x - 4) = 0 \] - \(x=0\) \(\Rightarrow y=0\) - \(x=4\) \(\Rightarrow y=4(4)=16\) The points of intersection are at \((0,0)\) and \((4,16)\). --- ### **Step 2: Set up the region** For \(x\) between 0 and 4, \( y \) varies between the lower curve \( y = x^2 \) and the upper line \( y=4x \). Thus, the region \( R \): \[ 0 \leq x \leq 4, \quad x^2 \leq y \leq 4x \] --- ### **Step 3: Write the double integral** \[ I = \int_{x=0}^{4} \int_{y=x^2}^{4x} xy \, dy \, dx \] --- ### **Step 4: Integrate with respect to \( y \)** \[ I = \int_{0}^{4} x \left[ \frac{1}{2} y^2 \right]_{y=x^2}^{y=4x} dx \] \[ = \int_{0}^{4} x \left( \frac{1}{2} (4x)^2 - \frac{1}{2} (x^2)^2 \right) dx \] Simplify: \[ = \int_{0}^{4} x \left( \frac{1}{2} \times 16 x^2 - \frac{1}{2} x^4 \right) dx \] \[ = \int_{0}^{4} x \left( 8 x^2 - \frac{1}{2} x^4 \right) dx \] Distribute \(x\): \[ = \int_{0}^{4} \left( 8 x^3 - \frac{1}{2} x^5 \right) dx \] --- ### **Step 5: Integrate with respect to \( x \)** \[ I = \left[ 8 \times \frac{x^4}{4} - \frac{1}{2} \times \frac{x^6}{6} \right]_0^4 \] Simplify: \[ = \left[ 2 x^4 - \frac{1}{12} x^6 \right]_0^4 \] Evaluate at \(x=4\): \[ = 2 \times (4)^4 - \frac{1}{12} \times (4)^6 \] Calculate: - \(4^4 = 256\) - \(4^6 = 4096\) Plug in: \[ = 2 \times 256 - \frac{1}{12} \times 4096 \] \[ = 512 - \frac{4096}{12} \] Simplify: \[ = 512 - \frac{1024}{3} \] Express both over a common denominator: \[ = \frac{1536}{3} - \frac{1024}{3} = \frac{1536 - 1024}{3} = \frac{512}{3} \] --- ## **Final Answer:** \[ \boxed{ I = \frac{512}{3} } \] --- Would you like to explore **Question 3** involving the triangle region or set notation?

Let's proceed with **Question 3**, which involves the triangle in \(\mathbb{R}^2\) with vertices \((1,1)\), \((2,3)\), and \((4,2)\). --- ## 3. (Sections 13.1-13.5) ### **Problem:** **Describe the triangle region \(A\)** with vertices: \[ (1,1), \quad (2,3), \quad (4,2) \] using set-builder notation of **Type I** and **Type II** regions. --- ### **Step 1: Find the equations of the sides** **Side 1:** between \((1,1)\) and \((2,3)\) - Slope: \[ m_{12} = \frac{3 - 1}{2 - 1} = \frac{2}{1} = 2 \] - Equation: \[ y - 1 = 2(x - 1) \Rightarrow y = 2x - 1 \] **Side 2:** between \((2,3)\) and \((4,2)\) - Slope: \[ m_{23} = \frac{2 - 3}{4 - 2} = \frac{-1}{2} = -\frac{1}{2} \] - Equation: \[ y - 3 = -\frac{1}{2}(x - 2) \Rightarrow y = -\frac{1}{2}x + 4 \] **Side 3:** between \((1,1)\) and \((4,2)\) - Slope: \[ m_{13} = \frac{2 - 1}{4 - 1} = \frac{1}{3} \] - Equation: \[ y - 1 = \frac{1}{3}(x - 1) \Rightarrow y = \frac{1}{3}x + \frac{2}{3} \] --- ### **Step 2: Determine the region** Plotting or analyzing these lines: - \(L_1\): \( y = 2x - 1 \) (passes through \((1,1)\)) - \(L_2\): \( y = -\frac{1}{2}x + 4 \) (passes through \((2,3)\)) - \(L_3\): \( y = \frac{1}{3}x + \frac{2}{3} \) (passes through \((1,1)\)) The vertices are at the intersections: - \(A = (1,1)\) - \(B = (2,3)\) - \(C = (4,2)\) --- ### **Step 3: Set-Builder Notation** #### **Type I (Vertical Strips):** For \(x\) in the interval between the minimum and maximum \(x\)-coordinate, i.e., from 1 to 4. At a fixed \(x\), \( y \) ranges between the lower and upper boundaries. - For \(x\) between 1 and 4: \[ \boxed{ A = \left\{ (x, y) \mid 1 \leq x \leq 4, \quad y_{\text{bottom}}(x) \leq y \leq y_{\text{top}}(x) \right\} } \] **Determine \( y_{\text{bottom}}(x) \) and \( y_{\text{top}}(x) \):** - For \(1 \leq x \leq 2\): The lower boundary is \( y = \frac{1}{3}x + \frac{2}{3} \) (line \(L_3\)) The upper boundary is \( y = 2x - 1 \) (line \(L_1\)) - For \(2 \leq x \leq 4\): The lower boundary is \( y = -\frac{1}{2}x + 4 \) (line \(L_2\)) The upper boundary is still \( y = 2x - 1 \) **Set-builder notation:** \[ A = \left\{ (x, y) \mid 1 \leq x \leq 2, \quad \frac{1}{3}x + \frac{2}{3} \leq y \leq 2x - 1 \right\} \cup \left\{ (x, y) \mid 2 \leq x \leq 4, \quad -\frac{1}{2}x + 4 \leq y \leq 2x - 1 \right\} \] --- #### **Type II (Horizontal Strips):** For fixed \( y \), describe the range of \( x \). - Find the \(x\)-limits for each \( y \) between the lowest and highest points. From the equations: - \( y = \frac{1}{3}x + \frac{2}{3} \Rightarrow x = 3(y - \frac{2}{3}) = 3y - 2 \) - \( y = 2x - 1 \Rightarrow x = \frac{y + 1}{2} \) - \( y = -\frac{1}{2}x + 4 \Rightarrow x = -2(y - 4) = -2 y + 8 \) Vertices: - \(A=(1,1)\): check \( y=1 \) - \(B=(2,3)\): check \( y=3 \) - \(C=(4,2)\): check \( y=2 \) Determine the \( y \)-interval: - The lowest \( y \)-value is at \( y=1 \), highest at \( y=3 \). At \( y \in [1, 3] \): - For \( y \) between 1 and 2: \( x \) runs between: \[ x = 3 y - 2 \quad (\text{left boundary}) \quad \text{and} \quad x = \frac{y+1}{2} \quad (\text{right boundary}) \] - For \( y \) between 2 and 3: \( x \) runs between: \[ x = 3 y - 2 \quad (\text{left boundary}) \quad \text{and} \quad x = -2 y + 8 \quad (\text{right boundary}) \] **Set-builder notation:** \[ A = \left\{ (x, y) \mid 1 \leq y \leq 2, \quad 3 y - 2 \leq x \leq \frac{y + 1}{2} \right\} \cup \left\{ (x, y) \mid 2 \leq y \leq 3, \quad 3 y - 2 \leq x \leq -2 y + 8 \right\} \] --- ## **Summary:** ### **Type I (Vertical slices):** \[ \boxed{ A = \left\{ (x, y) \mid 1 \leq x \leq 2, \quad \frac{1}{3}x + \frac{2}{3} \leq y \leq 2x - 1 \right\} \cup \left\{ (x, y) \mid 2 \leq x \leq 4, \quad -\frac{1}{2}x + 4 \leq y \leq 2x - 1 \right\} } \] ### **Type II (Horizontal slices):** \[ \boxed{ A = \left\{ (x, y) \mid 1 \leq y \leq 2, \quad 3 y - 2 \leq x \leq \frac{y + 1}{2} \right\} \cup \left\{ (x, y) \mid 2 \leq y \leq 3, \quad 3 y - 2 \leq x \leq -2 y + 8 \right\} } \] --- Would you like to continue with the last question involving the region of integration and its geometric interpretation?