The optimization problem in the image involves maximizing a Cobb-Douglas type function $x^\alpha y^\beta$, subject to a set of linear constraints. Here’s the step-by-step solution:

Problem

Maximize:
$z = x^\alpha y^\beta$
Subject to:
$x + 2y \leq 2, \quad x > 0, \quad y > 0, \quad \alpha > 0, \quad \beta > 0, \quad \alpha + \beta \leq 1.$

Step 1: Transform to Logarithmic Form

Taking the natural logarithm of the objective function: $\ln z = \alpha \ln x + \beta \ln y.$ Let $u = \ln z$, $v = \ln x$, and $w = \ln y$, so: $u = \alpha v + \beta w.$ This simplifies the problem to maximizing $u$ under the given constraints.

Step 2: Solve the Constraints

From the main constraint $x + 2y \leq 2$, substitute $y = \frac{2 - x}{2}$ for valid values of $x$. Note that $x > 0$ and $y > 0$, which means: $0 < x < 2.$

Step 3: Substitute into the Objective

The objective is $z = x^\alpha y^\beta$. Substituting $y = \frac{2 - x}{2}$: $z = x^\alpha \left(\frac{2 - x}{2}\right)^\beta.$

This can be written as: $z = \frac{x^\alpha (2 - x)^\beta}{2^\beta}.$

Step 4: Take the First-Order Condition

Maximize $z$ by taking the derivative with respect to $x$. Let: $f(x) = x^\alpha (2 - x)^\beta.$

The derivative is: $f'(x) = \alpha x^{\alpha - 1} (2 - x)^\beta - \beta x^\alpha (2 - x)^{\beta - 1}.$

Setting $f'(x) = 0$: $\alpha x^{\alpha - 1} (2 - x)^\beta = \beta x^\alpha (2 - x)^{\beta - 1}.$

Divide through by $x^{\alpha - 1} (2 - x)^{\beta - 1}$ (valid for $x > 0$ and $2 - x > 0$): $\alpha (2 - x) = \beta x.$

Rearrange: $\alpha (2 - x) = \beta x \quad \implies \quad 2\alpha = x(\alpha + \beta).$

Thus: $x = \frac{2\alpha}{\alpha + \beta}.$

Step 5: Solve for $y$

Using $x + 2y = 2$: $y = \frac{2 - x}{2} = \frac{2 - \frac{2\alpha}{\alpha + \beta}}{2}.$

Simplify: $y = \frac{2(\alpha + \beta) - 2\alpha}{2(\alpha + \beta)} = \frac{2\beta}{2(\alpha + \beta)} = \frac{\beta}{\alpha + \beta}.$

Step 6: Substitute into Objective

The maximum occurs at: $x = \frac{2\alpha}{\alpha + \beta}, \quad y = \frac{\beta}{\alpha + \beta}.$

Substitute these into $z = x^\alpha y^\beta$: $z = \left(\frac{2\alpha}{\alpha + \beta}\right)^\alpha \left(\frac{\beta}{\alpha + \beta}\right)^\beta.$

Simplify: $z = \frac{(2\alpha)^\alpha (\beta)^\beta}{(\alpha + \beta)^{\alpha + \beta}}.$

Final Answer

The maximum value of $z$ is: $z = \frac{(2\alpha)^\alpha (\beta)^\beta}{(\alpha + \beta)^{\alpha + \beta}},$ occurring at: $x = \frac{2\alpha}{\alpha + \beta}, \quad y = \frac{\beta}{\alpha + \beta}.$

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Related Questions

1. What happens when $\alpha + \beta = 1$?
2. Can this optimization problem be solved using the Lagrange multiplier method?
3. How does the solution change if $\alpha$ and $\beta$ are equal?
4. What is the economic interpretation of the Cobb-Douglas function in this context?
5. How does the result behave as $\beta \to 0$?

Tip

To better understand constrained optimization problems, practice solving them using both substitution and Lagrange multipliers to gain intuition for when each method is more appropriate.