Math Problem Statement

Given the optimization problem: Maximize -2x^2 - 2y^2 + xy subject to the constraint y - 3x = 16. Solve this using the Lagrange method.

Solution

This problem involves solving a multivariate optimization problem with constraints using the Lagrange method. Let’s break this down step by step.

Problem:

Maximize the function:
f(x,y)=2x22y2+xyf(x, y) = -2x^2 - 2y^2 + xy subject to the constraint:
g(x,y)=y3x16=0g(x, y) = y - 3x - 16 = 0

Step 1: Define the Lagrange Function

The Lagrange function is:
L(x,y,λ)=f(x,y)+λg(x,y)\mathcal{L}(x, y, \lambda) = f(x, y) + \lambda g(x, y) Substituting f(x,y)f(x, y) and g(x,y)g(x, y):
L(x,y,λ)=2x22y2+xy+λ(y3x16)\mathcal{L}(x, y, \lambda) = -2x^2 - 2y^2 + xy + \lambda (y - 3x - 16)

Step 2: Compute the Partial Derivatives

We compute the partial derivatives of L\mathcal{L} with respect to xx, yy, and λ\lambda, and set them to 0.

Partial derivative with respect to xx:

= -4x + y - 3\lambda$$ Set $$\frac{\partial \mathcal{L}}{\partial x} = 0$$: $$-4x + y - 3\lambda = 0 \quad \text{(1)}$$ #### Partial derivative with respect to $$y$$: $$\frac{\partial \mathcal{L}}{\partial y} = \frac{\partial}{\partial y}(-2x^2 - 2y^2 + xy + \lambda(y - 3x - 16)) = -4y + x + \lambda$$ Set $$\frac{\partial \mathcal{L}}{\partial y} = 0$$: $$-4y + x + \lambda = 0 \quad \text{(2)}$$ #### Partial derivative with respect to $$\lambda$$: $$\frac{\partial \mathcal{L}}{\partial \lambda} = y - 3x - 16$$ Set $$\frac{\partial \mathcal{L}}{\partial \lambda} = 0$$: $$y - 3x - 16 = 0 \quad \text{(3)}$$ ### Step 3: Solve the System of Equations We now solve the system of equations (1), (2), and (3): 1. $$-4x + y - 3\lambda = 0$$ 2. $$-4y + x + \lambda = 0$$ 3. $$y - 3x - 16 = 0$$ #### Substitute $$y = 3x + 16$$ (from equation 3) into equations (1) and (2): 1. $$-4x + (3x + 16) - 3\lambda = 0$$ $$-x + 16 - 3\lambda = 0 \quad \Rightarrow \quad \lambda = \frac{16 - x}{3} \quad \text{(4)}$$ 2. $$-4(3x + 16) + x + \lambda = 0$$ $$-12x - 64 + x + \lambda = 0 \quad \Rightarrow \quad -11x - 64 + \lambda = 0 \quad \text{(5)}$$ #### Substitute $$\lambda = \frac{16 - x}{3}$$ (from equation 4) into equation (5): $$-11x - 64 + \frac{16 - x}{3} = 0$$ Multiply through by 3 to eliminate the fraction: $$-33x - 192 + 16 - x = 0$$ $$-34x - 176 = 0$$ $$x = -\frac{176}{-34} = \frac{88}{17}$$ #### Solve for $$y$$: Using $$y = 3x + 16$$: $$y = 3\left(\frac{88}{17}\right) + 16 = \frac{264}{17} + \frac{272}{17} = \frac{536}{17}$$ #### Solve for $$\lambda$$: Using $$\lambda = \frac{16 - x}{3}$$: $$\lambda = \frac{16 - \frac{88}{17}}{3} = \frac{\frac{272}{17} - \frac{88}{17}}{3} = \frac{\frac{184}{17}}{3} = \frac{184}{51}$$ ### Step 4: Verify and Compute the Maximum Value Substitute $$x = \frac{88}{17}$$ and $$y = \frac{536}{17}$$ into $$f(x, y)$$: $$f\left(\frac{88}{17}, \frac{536}{17}\right) = -2\left(\frac{88}{17}\right)^2 - 2\left(\frac{536}{17}\right)^2 + \frac{88}{17} \cdot \frac{536}{17}$$ Would you like me to compute the exact numerical value of $$f(x, y)$$ for the maximum? ### Follow-up Questions: 1. What is the purpose of the Lagrange multiplier in optimization problems? 2. Why do we need to check second-order conditions to confirm maxima or minima? 3. How does the constraint $$y - 3x - 16 = 0$$ affect the optimization problem? 4. What would change if we were minimizing instead of maximizing the function? 5. Can this method be applied to problems with more than two variables? **Tip:** Always ensure your constraint is satisfied when solving optimization problems with Lagrange multipliers.

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Math Problem Analysis

Mathematical Concepts

Multivariate Calculus
Lagrange Multipliers
Optimization with Constraints

Formulas

Lagrange function: \( \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda g(x, y) \)
Partial derivatives: \( \frac{\partial \mathcal{L}}{\partial x}, \frac{\partial \mathcal{L}}{\partial y}, \frac{\partial \mathcal{L}}{\partial \lambda} \)

Theorems

Lagrange Multiplier Theorem

Suitable Grade Level

University Level (Advanced Calculus)