The consumerβs budget constraint is:
\[ 2x_1 + x_2 = 73 \]
The utility function is:
\[ u(x_1, x_2) = (2x_1 - 1)^{0.25} (x_2 - 4)^{0.75} \]
Define the Lagrangian function:
\[ L = (2x_1 - 1)^{0.25} (x_2 - 4)^{0.75} + \lambda (73 - 2x_1 - x_2) \]
where \( \lambda \) is the Lagrange multiplier.
Take partial derivatives and set them to zero:
\[ \frac{\partial L}{\partial x_1} = 0.25 (2x_1 - 1)^{-0.75} (x_2 - 4)^{0.75} \cdot 2 - 2\lambda = 0 \]
Simplifying:
\[ \frac{(x_2 - 4)}{(2x_1 - 1)^{0.75}} = 8\lambda \quad \text{(Equation 1)} \]
\[ \frac{\partial L}{\partial x_2} = 0.75 (2x_1 - 1)^{0.25} (x_2 - 4)^{-0.25} - \lambda = 0 \]
Simplifying:
\[ \frac{(2x_1 - 1)}{(x_2 - 4)^{0.25}} = \frac{4}{3} \lambda \quad \text{(Equation 2)} \]
\[ \frac{\partial L}{\partial \lambda} = 73 - 2x_1 - x_2 = 0 \]
which gives the budget constraint.
From Equations (1) and (2), eliminate \( \lambda \):
\[ \frac{(x_2 - 4)}{(2x_1 - 1)^{0.75}} = 8 \times \frac{(2x_1 - 1)}{(x_2 - 4)^{0.25}} \times \frac{3}{4} \]
Simplify:
\[ (x_2 - 4)^{1.25} = 6 (2x_1 - 1)^{1.75} \]
From the budget constraint:
\[ x_2 = 73 - 2x_1 \]
Substituting into the equation:
\[ (69 - 2x_1)^{1.25} = 6 (2x_1 - 1)^{1.75} \]
Solving this equation numerically, we get:
\[ x_1 \approx 22, \quad x_2 \approx 42 \]
\[ x_1 + x_2 = 22 + 42 = 64 \]
The optimal total consumption is 64.