Let's break down and solve the problem step by step: --- ### **Problem Restatement** Find the maximum and minimum values of \( f(x, y) \) in the region \( 4x^2 + 9y^2 \leq 36 \). --- ### **Step 1: Interpreting the Problem** - The region \( 4x^2 + 9y^2 \leq 36 \) describes an ellipse. - We are to maximize/minimize \( f(x, y) \) (which appears to be unspecified in the cropped image, but typically it's \( f(x, y) = x^2 + y^2 \) in such problems unless otherwise specified). Assuming \( f(x, y) = x^2 + y^2 \). --- ### **Step 2: Express the Constraint** \[ 4x^2 + 9y^2 \leq 36 \implies \frac{x^2}{9} + \frac{y^2}{4} \leq 1 \] This is an ellipse centered at the origin, with semi-axes \( a = 3 \) (x-axis), \( b = 2 \) (y-axis). --- ### **Step 3: Use Lagrange Multipliers** Set up the Lagrangian: \[ \mathcal{L}(x, y, \lambda) = x^2 + y^2 - \lambda (4x^2 + 9y^2 - 36) \] Take partial derivatives and set to zero: #### (1) \( \frac{\partial \mathcal{L}}{\partial x} = 0 \): \[ 2x - \lambda (8x) = 0 \implies x(1 - 4\lambda) = 0 \] #### (2) \( \frac{\partial \mathcal{L}}{\partial y} = 0 \): \[ 2y - \lambda (18y) = 0 \implies y(1 - 9\lambda) = 0 \] #### (3) \( \frac{\partial \mathcal{L}}{\partial \lambda} = 0 \): \[ 4x^2 + 9y^2 = 36 \] --- #### **Case 1: \( x = 0 \)** Plug into constraint: \[ 4(0)^2 + 9y^2 = 36 \implies y^2 = 4 \implies y = \pm 2 \] So, points: \( (0, 2) \) and \( (0, -2) \). \[ f(0, \pm 2) = 0^2 + (\pm 2)^2 = 4 \] --- #### **Case 2: \( y = 0 \)** Plug into constraint: \[ 4x^2 + 9(0)^2 = 36 \implies x^2 = 9 \implies x = \pm 3 \] So, points: \( (3, 0) \), \( (-3, 0) \). \[ f(\pm 3, 0) = (\pm 3)^2 + 0^2 = 9 \] --- #### **Case 3: \( 1 - 4\lambda = 0 \) and \( 1 - 9\lambda = 0 \)** Set \( 1 - 4\lambda = 0 \implies \lambda = \frac{1}{4} \) Set \( 1 - 9\lambda = 0 \implies \lambda = \frac{1}{9} \) These can't be satisfied simultaneously unless \( x = 0 \) or \( y = 0 \). --- ### **Step 4: Check Interior Point (0,0)** \( f(0,0) = 0 \), and \( 4(0)^2 + 9(0)^2 = 0 \leq 36 \). --- ### **Step 5: Compare Values** - At \( (0, 2) \) and \( (0, -2) \): \( f = 4 \) - At \( (3, 0) \) and \( (-3, 0) \): \( f = 9 \) - At \( (0, 0) \): \( f = 0 \) So, the **maximum** is at \( (\pm 3, 0) \) with \( f = 9 \). The **minimum** is at \( (0, 0) \) with \( f = 0 \). --- ## **Final Answers** **Maximum value:** \( \boxed{9} \) at \( (3, 0) \) and \( (-3, 0) \) **Minimum value:** \( \boxed{0} \) at \( (0, 0) \) --- If \( f(x, y) \) is something different, please specify!