To find the maxima and minima of $u = xy$, subject to the constraint $\frac{x}{2} + \frac{y}{3} = 1$, we use Lagrange multipliers.

Steps:

1. Define the objective function $u(x, y) = xy$ and the constraint $g(x, y) = \frac{x}{2} + \frac{y}{3} - 1 = 0$.
2. Set up the Lagrange function: $\mathcal{L}(x, y, \lambda) = xy + \lambda \left( \frac{x}{2} + \frac{y}{3} - 1 \right)$
3. Compute the partial derivatives of $\mathcal{L}$ and solve the system of equations: $\frac{\partial \mathcal{L}}{\partial x} = y + \lambda \frac{1}{2} = 0$ $\frac{\partial \mathcal{L}}{\partial y} = x + \lambda \frac{1}{3} = 0$ $\frac{\partial \mathcal{L}}{\partial \lambda} = \frac{x}{2} + \frac{y}{3} - 1 = 0$

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Step-by-step solution:

1. From $\frac{\partial \mathcal{L}}{\partial x} = 0$, we get: $y + \frac{\lambda}{2} = 0 \quad \Rightarrow \quad \lambda = -2y$

2. From $\frac{\partial \mathcal{L}}{\partial y} = 0$, we get: $x + \frac{\lambda}{3} = 0 \quad \Rightarrow \quad \lambda = -3x$

3. Equate the two expressions for $\lambda$: $-2y = -3x \quad \Rightarrow \quad \frac{y}{x} = \frac{3}{2} \quad \Rightarrow \quad y = \frac{3}{2}x$

4. Substitute $y = \frac{3}{2}x$ into the constraint equation: $\frac{x}{2} + \frac{\frac{3}{2}x}{3} = 1$ Simplify: $\frac{x}{2} + \frac{x}{2} = 1 \quad \Rightarrow \quad x = 1$

5. Using $x = 1$, substitute into $y = \frac{3}{2}x$: $y = \frac{3}{2}(1) = \frac{3}{2}$

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Results:

• The point is $(x, y) = (1, \frac{3}{2})$.
• The value of $u = xy = 1 \cdot \frac{3}{2} = \frac{3}{2}$.
• Since there is only one critical point and the constraint defines a boundary, this is the maximum value of $u$. The minimum is $u = 0$ when $x$ or $y$ is zero (but only on the boundary of the feasible region).

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Do you want me to provide a detailed graph or additional explanation?
Here are 5 related questions for further exploration:

1. Can we solve this problem using substitution instead of Lagrange multipliers?
2. What happens to $u$ if the constraint changes to $\frac{x}{2} + \frac{y}{3} \leq 1$?
3. How does the value of $u$ change if $x$ and $y$ are restricted to integers?
4. Can we apply Lagrange multipliers to maximize $u = x^2y$ under the same constraint?
5. What geometric insights can we gain from plotting $u = xy$ and the constraint?

Tip: Always ensure that all critical points satisfy the constraint before determining maxima or minima!