Let's solve the problem step by step: Given: \( f(x, y) = x^2 - y \) We're asked to find the minimum and maximum values of \( f \) on the triangle shown. ### **Step 1: Identify the Triangle Vertices** From the graph, the triangle has vertices at: - \( A(-1, 1) \) - \( B(1, 1) \) - \( C(0, -2) \) ### **Step 2: Check for Critical Points Inside the Triangle** Find the critical points by setting the gradient of \( f \) to zero: \[ \nabla f = (2x, -1) \] Setting to zero: \[ 2x = 0 \implies x = 0 \\ -1 = 0 \implies \text{No solution} \] So, there are **no critical points inside** the triangle. ### **Step 3: Check f on the Edges and Vertices** Let's evaluate \( f \) at the vertices: #### **At \( A(-1, 1) \):** \[ f(-1, 1) = (-1)^2 - 1 = 1 - 1 = 0 \] #### **At \( B(1, 1) \):** \[ f(1, 1) = (1)^2 - 1 = 1 - 1 = 0 \] #### **At \( C(0, -2) \):** \[ f(0, -2) = (0)^2 - (-2) = 0 + 2 = 2 \] ### **Step 4: Check the Edges** #### **Edge AB (\( y = 1, -1 \leq x \leq 1 \)):** \[ f(x, 1) = x^2 - 1 \] Range: \( x^2 \) is minimized at \( x = 0 \) (\( f(0, 1) = -1 \)), and maximized at \( x = \pm 1 \) (\( f(\pm 1, 1) = 0 \)). #### **Edge BC (\( y = -\frac{3}{2}x + \frac{3}{2}, 0 \leq x \leq 1 \)):** \[ y = -\frac{3}{2}x + \frac{3}{2} \] \[ f(x, y) = x^2 - \left( -\frac{3}{2}x + \frac{3}{2} \right ) = x^2 + \frac{3}{2}x - \frac{3}{2} \] This is a quadratic in \( x \) defined on \( [0, 1] \): - At \( x = 0: f(0, y) = 0 + 0 - \frac{3}{2} = -\frac{3}{2} \) - At \( x = 1: f(1, y) = 1 + \frac{3}{2} - \frac{3}{2} = 1 \) #### **Edge CA (\( y = \frac{3}{2}x + \frac{3}{2}, -1 \leq x \leq 0 \)):** \[ y = \frac{3}{2}x + \frac{3}{2} \] \[ f(x, y) = x^2 - \left( \frac{3}{2}x + \frac{3}{2} \right ) = x^2 - \frac{3}{2}x - \frac{3}{2} \] This is a quadratic in \( x \) defined on \( [-1, 0] \): - At \( x = -1: f(-1, y) = 1 + \frac{3}{2} - \frac{3}{2} = 1 \) - At \( x = 0: f(0, y) = 0 - 0 - \frac{3}{2} = -\frac{3}{2} \) ### **Step 5: Summary of Values** - At \( (0, 1) \): \( f = -1 \) - At \( (1, 1) \) or \( (-1, 1) \): \( f = 0 \) - At \( (0, -2) \): \( f = 2 \) - At \( (0, y) \) on \( y = 1 \) or \( y = -2 \): Check endpoints already done. - At \( (0, y) \) on slanted edges: At \( x = 0 \), both give \( f = -\frac{3}{2} \), which is less than -1. ### **Final Step: Minimum and Maximum** - **Minimum:** \( f(0, y) = -\frac{3}{2} \) at \( x = 0 \), on the edges \( BC \) and \( CA \) where \( y = \frac{3}{2} \) (which is not inside the triangle). But for the endpoints: - At \( x = 0, y = 1 \): \( f = -1 \) - At \( x = 0, y = -2 \): \( f = 2 \) - At \( x = 0, y = -\frac{3}{2} \): this is between the endpoints, so \( f = -\frac{3}{2} \) at \( (0, -\frac{3}{2}) \). But **(0, -3/2)** is not a vertex, but it is on the edge, so it is a valid minimum. - **Maximum:** At \( (0, -2) \), \( f = 2 \). --- ## **Final Answer** - **Maximum at \( (0, -2) \): \( f = 2 \)** - **Minimum at \( (0, -\frac{3}{2}) \): \( f = -\frac{3}{2} \)** --- ### **Label on the Triangle:** - Mark **maximum** at \( (0, -2) \). - Mark **minimum** at \( (0, -\frac{3}{2}) \) on the edge from \( (-1,1) \) to \( (0,-2) \) and from \( (1,1) \) to \( (0,-2) \). If you need the exact coordinates to mark on the triangle, **minimum** is at \( (0, -\frac{3}{2}) \), and **maximum** is at \( (0, -2) \).

Let's carefully analyze the problem and the given data. --- ### Given: Vertices of the triangle: - \( A(-1, 1) \) - \( B(-1, -1) \) - \( C(2, -1) \) Function: \[ f(x, y) = x^2 - y \] ### Objective: Find the points where \(f\) attains its **minimum** and **maximum** on the triangle with vertices as above. --- ### **Step 1: Visualize the Triangle** Vertices: - \( A(-1, 1) \) - \( B(-1, -1) \) - \( C(2, -1) \) The triangle has: - Left side: \( A \) to \( B \) (vertical line \( x = -1 \)) - Bottom side: \( B \) to \( C \) (horizontal line \( y = -1 \)) - Right side: \( A \) to \( C \) --- ### **Step 2: Check for critical points inside the triangle** Gradient of \( f \): \[ \nabla f = (2x, -1) \] Set equal to zero: \[ 2x = 0 \Rightarrow x=0 \] \[ -1=0 \Rightarrow \text{No solution} \] **Conclusion:** No critical points inside the triangle. --- ### **Step 3: Evaluate \(f\) at the vertices** - At \( A(-1, 1) \): \[ f(-1, 1) = (-1)^2 - 1 = 1 - 1 = 0 \] - At \( B(-1, -1) \): \[ f(-1, -1) = 1 - (-1) = 1 + 1 = 2 \] - At \( C(2, -1) \): \[ f(2, -1) = (2)^2 - (-1) = 4 + 1 = 5 \] --- ### **Step 4: Check \(f\) along the edges** --- **Edge \( AB \):** \( x = -1 \), \( y \) from 1 to -1 \[ f(-1, y) = (-1)^2 - y = 1 - y \] - At \( y=1 \): \( f=1 - 1=0 \) - At \( y=-1 \): \( f=1 - (-1)=2 \) **Range on \( AB \):** from 0 to 2. --- **Edge \( BC \):** \( y = -1 \), \( x \) from -1 to 2 \[ f(x, -1) = x^2 - (-1) = x^2 + 1 \] - At \( x=-1 \): \( 1 + 1=2 \) - At \( x=2 \): \( 4 + 1=5 \) **Range on \( BC \):** from 2 to 5. --- **Edge \( AC \):** Connects \( A(-1, 1) \) to \( C(2, -1) \) Equation of line \( AC \): \[ \text{Slope} = \frac{-1 - 1}{2 - (-1)} = \frac{-2}{3} = -\frac{2}{3} \] Equation: \[ y - 1 = -\frac{2}{3}(x + 1) \] \[ y = 1 - \frac{2}{3}(x+1) = 1 - \frac{2}{3}x - \frac{2}{3} = \frac{1}{3} - \frac{2}{3}x \] Express \(f\) along \( AC \): \[ f(x, y) = x^2 - y = x^2 - \left(\frac{1}{3} - \frac{2}{3}x\right) = x^2 - \frac{1}{3} + \frac{2}{3}x \] Simplify: \[ f(x) = x^2 + \frac{2}{3}x - \frac{1}{3} \] Domain: - \( x \) from \( -1 \) to \( 2 \) Find extrema on \( x \in [-1, 2] \): Derivative: \[ f'(x) = 2x + \frac{2}{3} \] Set to zero: \[ 2x + \frac{2}{3} = 0 \Rightarrow 2x = -\frac{2}{3} \Rightarrow x = -\frac{1}{3} \] Check whether \( x=-\frac{1}{3} \) is within the domain: Yes, since \( -1 < -\frac{1}{3} < 2 \). Evaluate \(f\) at critical point: \[ f(-\frac{1}{3}) = \left(-\frac{1}{3}\right)^2 + \frac{2}{3} \times \left(-\frac{1}{3}\right) - \frac{1}{3} = \frac{1}{9} - \frac{2}{9} - \frac{1}{3} \] \[ = \frac{1}{9} - \frac{2}{9} - \frac{3}{9} = \frac{1 - 2 - 3}{9} = -\frac{4}{9} \] Now evaluate at endpoints: - At \( x=-1 \): \[ f(-1) = 1 + \frac{2}{3}(-1) - \frac{1}{3} = 1 - \frac{2}{3} - \frac{1}{3} = 1 - 1 = 0 \] - At \( x=2 \): \[ f(2) = 4 + \frac{2}{3} \times 2 - \frac{1}{3} = 4 + \frac{4}{3} - \frac{1}{3} = 4 + 1 = 5 \] --- ### **Step 5: Summary of \(f\) values** | Point / Segment | \(f\) value | Remarks | |------------------|--------------|---------| | \(A(-1,1)\) | 0 | Vertex | | \(B(-1,-1)\) | 2 | Vertex | | \(C(2,-1)\) | 5 | Vertex | | Along \(AB\): \( y \in [ -1, 1 ] \) | \(1 - y\) | Min at \( y=1 \), \(f=0\); Max at \( y=-1 \), \(f=2\) | | Along \(BC\): \( y=-1 \) | \(x^2 + 1\) | Min at \(x=-1\), \(f=2\); Max at \(x=2\), \(f=5\) | | Along \(AC\): \( x \in [-1, 2] \) | \(x^2 + \frac{2}{3}x - \frac{1}{3}\) | Min at \(x=-\frac{1}{3}\), \(f=-\frac{4}{9}\); endpoints \(x=-1\) and \(x=2\) give \(f=0\) and 5 | --- ### **Final Results:** - **Minimum value of \(f\):** \(\boxed{-\frac{4}{9}}\) at \( x = -\frac{1}{3} \) - **Corresponding \( y \):** \[ y = \frac{1}{3} - \frac{2}{3} \times \left(-\frac{1}{3}\right) = \frac{1}{3} + \frac{2}{9} = \frac{3}{9} + \frac{2}{9} = \frac{5}{9} \] So, the **minimum point** is approximately: \[ \boxed{\left( -\frac{1}{3}, \frac{5}{9} \right)} \] - **Maximum value of \(f\):** 5 at \( C(2, -1) \) --- ### **Final answer:** - **Maximum of \(f\):** **5** at **\( (2, -1) \)** - **Minimum of \(f\):** **\(-\frac{4}{9}\)** at approximately \( \left( -\frac{1}{3}, \frac{5}{9} \right) \) --- If marking on the triangle, label the point \(\left( -\frac{1}{3}, \frac{5}{9} \right)\) as the **minimum**, and vertex \( C(2, -1) \) as the **maximum**.