Let's solve the problem **step-by-step**: --- ### **Given:** - Function: \( f(x, y) = x^2 + 2x - y \) - Region \( R = \{ (x, y) \mid -1 \leq x \leq 1, \ 1 \leq y \leq 2 - x^2 \} \) We are to **find the extreme values (maximum and minimum) and the corresponding points**. --- ## **Step 1: Find critical points inside the region** First, find points where the gradient of \( f \) vanishes **inside** \( R \): \[ f(x, y) = x^2 + 2x - y \] \[ f_x = 2x + 2, \quad f_y = -1 \] Set the partial derivatives to zero: - \( f_x = 0 \implies 2x + 2 = 0 \implies x = -1 \) - \( f_y = 0 \implies -1 = 0 \) (**Never true!**) So, the only interior critical point is at \( x = -1 \). But what about \( y \)? At \( x = -1 \), \( y \) can be any value in \( [1, 2 - x^2] = [1, 2 - 1] = [1, 1] \), so only \( y = 1 \) is possible. So, the only possible interior critical point is at \( (-1, 1) \). --- ## **Step 2: Find extreme values on the boundary** The boundary of \( R \) consists of: 1. **Left edge:** \( x = -1, \ 1 \leq y \leq 1 \)               (so only the point \( (-1, 1) \)) 2. **Right edge:** \( x = 1, \ 1 \leq y \leq 1 \)                (so only the point \( (1, 1) \)) 3. **Bottom edge:** \( -1 \leq x \leq 1, \ y = 1 \) 4. **Top edge:** \( -1 \leq x \leq 1, \ y = 2 - x^2 \) Let's examine each: --- ### **(i) Left edge: \( x = -1, y = 1 \)** - \( f(-1, 1) = (-1)^2 + 2(-1) - 1 = 1 - 2 - 1 = -2 \) ### **(ii) Right edge: \( x = 1, y = 1 \)** - \( f(1, 1) = (1)^2 + 2(1) - 1 = 1 + 2 - 1 = 2 \) --- ### **(iii) Bottom edge: \( y = 1, -1 \leq x \leq 1 \)** Let’s write \( f(x, 1) = x^2 + 2x - 1 \), for \( -1 \leq x \leq 1 \). This is a quadratic in \( x \). Let's find its extreme values in this interval. - **Derivative:** \( \frac{d}{dx}(x^2 + 2x - 1) = 2x + 2 \) - Set \( 2x + 2 = 0 \implies x = -1 \), which is at the left endpoint. So the max/min will be at endpoints: - At \( x = -1 \): \( f(-1, 1) = -2 \) (already calculated) - At \( x = 1 \): \( f(1, 1) = 2 \) (already calculated) Thus, on the bottom edge, the function attains values from \( -2 \) (at \( x = -1 \)) to \( 2 \) (at \( x = 1 \)). --- ### **(iv) Top edge: \( y = 2 - x^2, -1 \leq x \leq 1 \)** Substitute \( y \): \[ f(x, 2 - x^2) = x^2 + 2x - (2 - x^2) = x^2 + 2x - 2 + x^2 = 2x^2 + 2x - 2 \] Find the extreme values of \( g(x) = 2x^2 + 2x - 2 \) on \( -1 \leq x \leq 1 \): - Derivative: \( g'(x) = 4x + 2 \) - Set \( 4x + 2 = 0 \implies x = -\frac{1}{2} \) Check at \( x = -1, x = 1, x = -\frac{1}{2} \): - At \( x = -1 \): \( y = 2 - (-1)^2 = 1 \), \( f(-1, 1) = -2 \) - At \( x = 1 \): \( y = 2 - (1)^2 = 1 \), \( f(1, 1) = 2 \) - At \( x = -\frac{1}{2} \): \( y = 2 - \left(-\frac{1}{2}\right)^2 = 2 - \frac{1}{4} = \frac{7}{4} \) \( f\left(-\frac{1}{2}, \frac{7}{4}\right) = \left(-\frac{1}{2}\right)^2 + 2 \cdot \left(-\frac{1}{2}\right) - \frac{7}{4} \) \[ = \frac{1}{4} - 1 - \frac{7}{4} = \frac{1}{4} - \frac{4}{4} - \frac{7}{4} = \frac{1 - 4 - 7}{4} = \frac{-10}{4} = -2.5 \] --- ## **Step 3: Find the maximum and minimum values** Let’s collect all candidate points: - \( (-1, 1) \)                \( f = -2 \) - \( (1, 1) \)                 \( f = 2 \) - \( ( -\frac{1}{2}, 1 ) \)     \( f = (-\frac{1}{2})^2 + 2(-\frac{1}{2}) - 1 = \frac{1}{4} - 1 - 1 = -1.75 \) - \( ( -\frac{1}{2}, \frac{7}{4} ) \) \( f = -2.5 \) Compare all values: | Point | Value | |----------------------------------|-----------| | \( (-1, 1) \) | \( -2 \) | | \( (1, 1) \) | \( 2 \) | | \( ( -\frac{1}{2}, 1 ) \) | \( -1.75 \) | | \( ( -\frac{1}{2}, \frac{7}{4} )\) | \( -2.5 \) | The minimum is at \( ( -\frac{1}{2}, \frac{7}{4} ) \) with value \( -2.5 \). The maximum is at \( (1, 1) \) with value \( 2 \). --- ## **Step 4: Final Answer** ### **Maximum value:** \[ \boxed{2} \text{ at } (1, 1) \] ### **Minimum value:** \[ \boxed{ -2.5 } \text{ at } \left( -\frac{1}{2}, \frac{7}{4} \right) \] --- ### **Summary Table** | Point | \( f(x, y) \) | |--------------------------------------------|---------------| | \( (-1, 1) \) | \( -2 \) | | \( (1, 1) \) | \( 2 \) | | \( \left(-\frac{1}{2}, \frac{7}{4}\right) \) | \( -2.5 \) | --- ## **Final Boxed Answers** \[ \boxed{ \begin{array}{l} \text{Maximum value is } 2 \text{ at } (1, 1) \\ \text{Minimum value is } -2.5 \text{ at } \left( -\frac{1}{2}, \frac{7}{4} \right) \end{array} } \]