Calculer les dérivées partielles de la fonction par rapport à Xᵢ, i=1,2

The given equations (1) and (2) are:

$$\alpha X_1^{\alpha-1} X_2^\beta = \lambda P_1 \quad (1)$$ $$\beta X_1^\alpha X_2^{\beta-1} = \lambda P_2 \quad (2)$$

Dividing (1) by (2):

$$\frac{\alpha X_1^{\alpha-1} X_2^\beta}{\beta X_1^\alpha X_2^{\beta-1}} = \frac{\lambda P_1}{\lambda P_2}$$ $$\frac{\alpha}{\beta} \cdot \frac{X_1^{\alpha-1}}{X_1^\alpha} \cdot \frac{X_2^\beta}{X_2^{\beta-1}} = \frac{P_1}{P_2}$$

Using exponent rules $a^m/a^n = a^{m-n}$:

$$\frac{\alpha}{\beta} \cdot X_1^{\alpha-1-\alpha} \cdot X_2^{\beta-(\beta-1)} = \frac{P_1}{P_2}$$ $$\frac{\alpha}{\beta} \cdot X_1^{-1} \cdot X_2^1 = \frac{P_1}{P_2}$$ $$\frac{\alpha}{\beta} \cdot \frac{X_2}{X_1} = \frac{P_1}{P_2}$$

$$X_2 = \frac{\beta}{\alpha} \cdot \frac{P_1}{P_2} \cdot X_1 \quad (4)$$

This matches the student's derived equation (4).

Equation (3) is given as:

$$P_1 X_1 + P_2 X_2 = R \quad (3)$$

Substitute the expression for $X_2$ from Step 3:

$$P_1 X_1 + P_2 \left( \frac{\beta}{\alpha} \frac{P_1}{P_2} X_1 \right) = R$$

$$P_1 X_1 + \frac{\beta}{\alpha} P_1 X_1 = R$$

Factor out $P_1 X_1$:

$$P_1 X_1 \left( 1 + \frac{\beta}{\alpha} \right) = R$$ $$P_1 X_1 \left( \frac{\alpha + \beta}{\alpha} \right) = R$$ $$X_1 = \frac{R \alpha}{P_1 (\alpha + \beta)}$$

So, the expression for $X_1$ is:

$$\boxed{X_1 = \frac{\alpha R}{P_1 (\alpha + \beta)}}$$

$$X_2 = \frac{\beta}{\alpha} \frac{P_1}{P_2} X_1$$ $$X_2 = \frac{\beta}{\alpha} \frac{P_1}{P_2} \left( \frac{\alpha R}{P_1 (\alpha + \beta)} \right)$$

$$X_2 = \frac{\beta}{\alpha} \frac{P_1}{P_2} \frac{\alpha R}{P_1 (\alpha + \beta)}$$

Cancel out $\alpha$ and $P_1$:

$$X_2 = \frac{\beta R}{P_2 (\alpha + \beta)}$$

So, the expression for $X_2$ is:

$$\boxed{X_2 = \frac{\beta R}{P_2 (\alpha + \beta)}}$$