Let's solve each question step-by-step: --- ### **1. Critical Points and Their Nature** Given: \[ f(x, y) = \frac{y}{x} - \ln x + \frac{1}{2} y^2 \] #### **Step 1: Find the critical points** Critical points occur where the gradient is zero: \[ \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (0, 0) \] Compute the partial derivatives: \[ \frac{\partial f}{\partial x} = -\frac{y}{x^2} - \frac{1}{x} \] \[ \frac{\partial f}{\partial y} = \frac{1}{x} + y \] Set both to zero: 1. \(-\frac{y}{x^2} - \frac{1}{x} = 0\) 2. \(\frac{1}{x} + y = 0\) From (2): \(y = -\frac{1}{x}\) Plug into (1): \[ -\frac{-1/x}{x^2} - \frac{1}{x} = 0 \implies \frac{1}{x^3} - \frac{1}{x} = 0 \] \[ \frac{1 - x^2}{x^3} = 0 \implies x^2 = 1 \implies x = \pm 1 \] So, - If \(x = 1\), \(y = -1\) - If \(x = -1\), \(y = 1\) **Critical points:** \((1, -1)\) and \((-1, 1)\) --- #### **Step 2: Nature of Critical Points (Second Derivative Test)** Compute second partials: \[ f_{xx} = \frac{2y}{x^3} + \frac{1}{x^2} \] \[ f_{yy} = 1 \] \[ f_{xy} = -\frac{1}{x^2} \] **At \((1, -1)\):** - \(f_{xx} = \frac{2(-1)}{1^3} + 1 = -2 + 1 = -1\) - \(f_{yy} = 1\) - \(f_{xy} = -1\) Hessian determinant: \[ D = f_{xx} f_{yy} - (f_{xy})^2 = (-1)(1) - (-1)^2 = -1 - 1 = -2 \] Since \(D < 0\), **saddle point** at \((1, -1)\). --- **At \((-1, 1)\):** - \(f_{xx} = \frac{2(1)}{(-1)^3} + \frac{1}{(-1)^2} = -2 + 1 = -1\) - \(f_{yy} = 1\) - \(f_{xy} = -1\) Hessian determinant: \[ D = (-1)(1) - (-1)^2 = -1 - 1 = -2 \] Since \(D < 0\), **saddle point** at \((-1, 1)\). --- ### **Final Answers for 1:** **Critical points:** \((1, -1)\) and \((-1, 1)\) **Nature:** Saddle points. --- ## **2. Maximum Volume with Lagrange Multipliers** **Let** - Length = \(l\) - Width = \(w\) - Height = \(h\) **Constraint:** \(l + 2w + h = 300\) **Objective:** Maximize \(V = lwh\) Let \(F = lwh\), constraint \(g(l, w, h) = l + 2w + h - 300 = 0\). Set up Lagrange: \[ \nabla F = \lambda \nabla g \] \[ \left( wh, lh, lw \right) = \lambda (1, 2, 1) \] So: 1. \(wh = \lambda\) 2. \(lh = 2\lambda\) 3. \(lw = \lambda\) From (1) and (3): \[ wh = lw \implies h = l \] From (1) and (2): \[ wh = lh/2 \implies w = l/2 \] So, \(h = l\), \(w = l/2\). Plug into constraint: \[ l + 2(w) + h = 300 \] \[ l + 2(l/2) + l = 300 \] \[ l + l + l = 300 \implies 3l = 300 \implies l = 100 \] So: - \(l = 100\) - \(w = 50\) - \(h = 100\) **Max Volume:** \[ V = lwh = 100 \times 50 \times 100 = \boxed{500,000 \text{ cm}^3} \] --- ### **3. Triangle Region and Integral** #### **(a) Sketch and Vertices** Vertices: (1, 1), (2, 3), (4, 2). #### **(b) Type I Region?** A Type I region is bounded as \(a \leq x \leq b\), \(g_1(x) \leq y \leq g_2(x)\). Find equations for the sides: 1. **(1,1) to (2,3):** Slope = 2, so \(y - 1 = 2(x - 1) \implies y = 2x - 1\) 2. **(2,3) to (4,2):** Slope = -0.5, so \(y - 3 = -0.5(x - 2) \implies y = -0.5x + 4\) 3. **(1,1) to (4,2):** Slope = 1/3, so \(y - 1 = \frac{1}{3}(x - 1) \implies y = \frac{1}{3}x + \frac{2}{3}\) For \(x\) in [1, 2], lower bound is \(y = 2x - 1\), upper is \(y = \frac{1}{3}x + \frac{2}{3}\). For \(x\) in [2, 4], lower bound is \(y = -0.5x + 4\), upper is \(y = \frac{1}{3}x + \frac{2}{3}\). So, region is a union of two Type I regions. **Set builder notation:** \[ R = \left\{ \begin{aligned} & (x, y) : 1 \leq x \leq 2,\, 2x-1 \leq y \leq \frac{1}{3}x + \frac{2}{3} \\ & \cup \\ & (x, y) : 2 \leq x \leq 4,\, -0.5x+4 \leq y \leq \frac{1}{3}x + \frac{2}{3} \end{aligned} \right\} \] #### **(c) Type II Region?** Type II: \(c \leq y \leq d\), \(h_1(y) \leq x \leq h_2(y)\). Find y-range: lowest \(y = 1\), highest \(y = 3\). For each y, solve for x in the triangle: 1. From \(y = 2x - 1\): \(x = \frac{y+1}{2}\) 2. From \(y = -0.5x + 4\): \(x = 2(4-y)\) 3. From \(y = \frac{1}{3}x + \frac{2}{3}\): \(x = 3y - 2\) Check intervals: - For \(1 \leq y \leq 2\): left: \(x = \frac{y+1}{2}\), right: \(x = 3y-2\) - For \(2 \leq y \leq 3\): left: \(x = 3y-2\), right: \(x = 2(4-y)\) So, union of two Type II regions. **Set builder notation:** \[ R = \left\{ \begin{aligned} & (x, y) : 1 \leq y \leq 2,\, \frac{y+1}{2} \leq x \leq 3y - 2 \\ & \cup \\ & (x, y) : 2 \leq y \leq 3,\, 3y - 2 \leq x \leq 8 - 2y \end{aligned} \right\} \] --- #### **(d) Integral \(\iint_R (x + y) dA\)** i. **Order \(dy\,dx\):** \[ \int_{x=1}^{2} \int_{y=2x-1}^{\frac{1}{3}x+\frac{2}{3}} (x+y) dy dx + \int_{x=2}^{4} \int_{y=-0.5x+4}^{\frac{1}{3}x+\frac{2}{3}} (x+y) dy dx \] ii. **Order \(dx\,dy\):** \[ \int_{y=1}^{2} \int_{x=\frac{y+1}{2}}^{3y-2} (x+y) dx dy + \int_{y=2}^{3} \int_{x=3y-2}^{8-2y} (x+y) dx dy \] --- ### **Summary of Answers** **1.** - Critical points: \((1, -1)\), \((-1, 1)\) - Both are saddle points. **2.** - Maximum volume: \(\boxed{500,000~\text{cm}^3}\) at \(l = 100\), \(w = 50\), \(h = 100\). **3.** - (a) Triangle with vertices (1,1), (2,3), (4,2). - (b) Union of two Type I regions (set notation above). - (c) Union of two Type II regions (set notation above). - (d) Double integrals set up as above (explicit limits). If you need full calculations for the integral, let me know!

Let's carefully proceed with each part of the question, providing detailed steps, explanations, and final answers. --- ### **Part 1: Critical Points and Their Nature** Given: \[ f(x, y) = \frac{y}{x} - \ln x + \frac{1}{2} y^2 \] --- #### **Step 1: Find the first partial derivatives** Compute: \[ \frac{\partial f}{\partial x} \quad \text{and} \quad \frac{\partial f}{\partial y} \] - \(\frac{\partial}{\partial x}\left(\frac{y}{x}\right) = -\frac{y}{x^2}\) - \(\frac{\partial}{\partial x}(- \ln x) = - \frac{1}{x}\) - \(\frac{\partial}{\partial x}\left(\frac{1}{2} y^2\right) = 0\) Thus: \[ f_x = -\frac{y}{x^2} - \frac{1}{x} \] Similarly: \[ f_y = \frac{\partial}{\partial y}\left(\frac{y}{x}\right) + \frac{\partial}{\partial y}(-\ln x) + \frac{\partial}{\partial y}\left(\frac{1}{2} y^2\right) = \frac{1}{x} + 0 + y = \frac{1}{x} + y \] --- #### **Step 2: Set derivatives to zero to find critical points** Solve: \[ f_x = 0 \implies -\frac{y}{x^2} - \frac{1}{x} = 0 \] \[ f_y = 0 \implies \frac{1}{x} + y = 0 \] From the second: \[ y = -\frac{1}{x} \] Substitute into the first: \[ -\frac{-1/x}{x^2} - \frac{1}{x} = 0 \] \[ \frac{1/x}{x^2} - \frac{1}{x} = 0 \] \[ \frac{1}{x^3} - \frac{1}{x} = 0 \] Multiply through by \(x^3\) (assuming \(x \neq 0\)): \[ 1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1 \] Corresponding \(y\): - For \(x=1\): \[ y = -\frac{1}{1} = -1 \] - For \(x=-1\): \[ y = -\frac{1}{-1} = 1 \] **Critical points:** \[ (1, -1) \quad \text{and} \quad (-1, 1) \] --- #### **Step 3: Determine the nature of these critical points** Compute second derivatives: \[ f_{xx} = \frac{\partial}{\partial x} \left(-\frac{y}{x^2} - \frac{1}{x}\right) \] - \(\frac{\partial}{\partial x} \left(-\frac{y}{x^2}\right) = 2 \frac{y}{x^3}\) - \(\frac{\partial}{\partial x} \left(- \frac{1}{x}\right) = \frac{1}{x^2}\) So: \[ f_{xx} = 2 \frac{y}{x^3} + \frac{1}{x^2} \] \[ f_{yy} = \frac{\partial}{\partial y} \left(\frac{1}{x} + y\right) = 1 \] \[ f_{xy} = \frac{\partial}{\partial y} \left(-\frac{y}{x^2} - \frac{1}{x}\right) = - \frac{1}{x^2} \] Now evaluate at each critical point. --- ### **At \((1, -1)\):** \[ f_{xx} = 2 \frac{-1}{1^3} + \frac{1}{1^2} = -2 + 1 = -1 \] \[ f_{yy} = 1 \] \[ f_{xy} = - \frac{1}{1^2} = -1 \] Hessian determinant: \[ D = f_{xx} f_{yy} - (f_{xy})^2 = (-1)(1) - (-1)^2 = -1 - 1 = -2 < 0 \] **Conclusion:** Saddle point at \((1, -1)\). --- ### **At \((-1, 1)\):** \[ f_{xx} = 2 \frac{1}{(-1)^3} + \frac{1}{(-1)^2} = 2 \times (-1) + 1 = -2 + 1 = -1 \] \[ f_{yy} = 1 \] \[ f_{xy} = - \frac{1}{(-1)^2} = -1 \] Hessian: \[ D = (-1)(1) - (-1)^2 = -1 - 1 = -2 < 0 \] **Conclusion:** Saddle point at \((-1, 1)\). --- ### **Final answer for Part 1:** - **Critical points:** \(\boxed{(1, -1) \text{ and } (-1, 1)}\) - **Nature:** Both are **saddle points**. --- --- ### **Part 2: Maximize Volume of a Rectangular Box** Given: \[ \text{Constraint:} \quad l + 2w + h = 300 \] \[ \text{Objective:} \quad V = l \times w \times h \] --- #### **Step 1: Use Lagrange multipliers** Set: \[ F(l,w,h) = lwh \] \[ g(l,w,h) = l + 2w + h - 300 = 0 \] Find \(\nabla F\) and \(\nabla g\): \[ \nabla F = (wh,\, l h,\, l w) \] \[ \nabla g = (1,\, 2,\, 1) \] Set: \[ \nabla F = \lambda \nabla g \] Thus: \[ wh = \lambda \quad (1) \] \[ l h = 2 \lambda \quad (2) \] \[ l w = \lambda \quad (3) \] --- #### **Step 2: Find relationships between \(l, w, h\)** From (1) and (3): \[ wh = l w \implies h = l \] (assuming \(w \neq 0\)) From (1) and (2): \[ wh = 2 \lambda \] But \(wh = \lambda\), so: \[ \lambda = 2 \lambda \implies \lambda = 0 \] But \(\lambda = 0\) implies \(wh=0\), which would mean volume zero, not a maximum. Alternatively, note that: From (1): \[ wh = \lambda \] From (3): \[ l w = \lambda \] Dividing (3) by (1): \[ \frac{l w}{w h} = \frac{\lambda}{\lambda} \implies \frac{l}{h} = 1 \implies l = h \] Similarly, from (2) and (1): \[ l h = 2 \lambda \] and \[ w h = \lambda \] Express \(\lambda\) from the second: \[ \lambda = w h \] and from the first: \[ l h = 2 w h \implies l = 2w \] but earlier we established \(l = h\), so: \[ h = l = 2w \] Now, use the constraint: \[ l + 2w + h = 300 \] Substitute \(l = 2w\), \(h= 2w\): \[ 2w + 2w + 2w = 300 \] \[ 6w = 300 \Rightarrow w = 50 \] \[ l = 2w = 100 \] \[ h= 2w= 100 \] --- #### **Step 3: Compute maximum volume** \[ V = l w h = 100 \times 50 \times 100 = \boxed{500,000 \text{ cm}^3} \] --- ### **Final answer for Part 2:** \[ \boxed{ \text{Maximum volume} = \boxed{500,000 \text{ cm}^3} } \] achieved when \(l=100\), \(w=50\), \(h=100\). --- --- ### **Part 3: Region R and Its Properties; Computing the Integral** Vertices: \((1, 1)\), \((2, 3)\), \((4, 2)\). --- #### **(a) Sketch and intersection points** - **Lines:** 1. From \((1, 1)\) to \((2, 3)\): \[ \text{Slope} = \frac{3 - 1}{2 - 1} = 2 \] Equation: \[ y - 1 = 2(x - 1) \implies y = 2x - 1 \] 2. From \((2,3)\) to \((4,2)\): \[ \text{Slope} = \frac{2 - 3}{4 - 2} = -\frac{1}{2} \] Equation: \[ y - 3 = -\frac{1}{2}(x - 2) \Rightarrow y = -\frac{1}{2}x + 4 \] 3. From \((1, 1)\) to \((4, 2)\): \[ \text{Slope} = \frac{2 - 1}{4 - 1} = \frac{1}{3} \] Equation: \[ y - 1 = \frac{1}{3}(x - 1) \Rightarrow y = \frac{1}{3}x + \frac{2}{3} \] --- #### **(b) Is \(R\) a Type I region?** - The region is bounded between: \[ x = 1 \quad \text{and} \quad x=4 \] - For \(x \in [1, 2]\): \[ \text{Lower boundary: } y = 2x - 1 \] \[ \text{Upper boundary: } y = \frac{1}{3}x + \frac{2}{3} \] - For \(x \in [2, 4]\): \[ \text{Lower boundary: } y = -\frac{1}{2}x + 4 \] \[ \text{Upper boundary: } y = \frac{1}{3}x + \frac{2}{3} \] Since the region can be expressed as: \[ R = \left\{ (x, y): 1 \leq x \leq 2, \quad 2x - 1 \leq y \leq \frac{1}{3}x + \frac{2}{3} \right\} \cup \left\{ (x, y): 2 \leq x \leq 4, \quad -\frac{1}{2}x + 4 \leq y \leq \frac{1}{3}x + \frac{2}{3} \right\} \] **Conclusion:** \(R\) is a union of two Type I regions. --- #### **(c) Is \(R\) a Type II region?** - The vertical extent: \[ y \in [1, 3] \] - For fixed \(y\), find \(x\)-limits: From the boundary lines: - \(y = 2x - 1 \Rightarrow x = \frac{y+1}{2}\) - \(y = -\frac{1}{2}x + 4 \Rightarrow x = 2(4 - y) = 8 - 2y\) - \(y = \frac{1}{3}x + \frac{2}{3} \Rightarrow x = 3(y - \frac{2}{3}) = 3y - 2\) - For \(y \in [1,2]\): \[ x \in \left[\frac{y+1}{2}, 3 y - 2 \right] \] - For \(y \in [2,3]\): \[ x \in \left[3 y - 2, 8 - 2 y \right] \] Thus, \(R\) can be written as a union of two Type II regions: \[ R = \left\{ ( x, y): 1 \leq y \leq 2, \quad \frac{y+1}{2} \leq x \leq 3 y - 2 \right\} \cup \left\{ ( x, y): 2 \leq y \leq 3, \quad 3 y - 2 \leq x \leq 8 - 2