Question

maximize the utility function u=x1+x2 subject to y=p1x1+p2x2 also find out demand function.

Answer 1

Answer on Question #58326 – Math – Other

Question

Maximize the utility function $u = x1 + x2$ subject to $y = p1x1 + p2x2$ also find out demand function.

Solution

Using the Lagrange multiplier obtain

$$L = u(x_1, x_2) + \lambda(y - p_1 x_1 - p_2 x_2)$$ $$\frac{\partial L}{\partial x_1} = 0,$$ $$\frac{\partial L}{\partial x_2} = 0,$$ $$\frac{\partial L}{\partial \lambda} = 0.$$

Hence,

$$\frac{\partial u}{\partial x_1} - \lambda p_1 = 0,$$ $$\frac{\partial u}{\partial x_2} - \lambda p_2 = 0,$$ $$p_1 x_1 + p_2 x_2 = y.$$

In this problem,

$$\frac{\partial}{\partial x_1}(x_1 + x_2) - \lambda p_1 = 0,$$ $$\frac{\partial}{\partial x_2}(x_1 + x_2) - \lambda p_2 = 0,$$ $$p_1 x_1 + p_2 x_2 = y,$$

which is equivalent to

$$\begin{array}{l} 1 - \lambda p _ {1} = 0, \\ 1 - \lambda p _ {2} = 0, \\ p _ {1} x _ {1} + p _ {2} x _ {2} = y \\ \end{array}$$

If $p_1 \neq p_2$ then there is no solution.

If $p_1 = p_2$ then $p_1(x_1 + x_2) = y$, therefore, $u^* = x_1 + x_2 = \frac{y}{p_1}$ is a maximum value of the utility function, here $x_1$ is any number satisfying $0 \leq x_1 \leq \frac{y}{p_1}$ and $x_2 = \frac{y}{p_1} - x_1$.

The solutions for $x_{1}$ and $x_{2}$ are called the consumer's demand functions.

**Remark.** It must be $u = x_{1}x_{2}$ because linear function does not have maximum.

$$y = p _ {1} x _ {1} + p _ {2} x _ {2} \rightarrow x _ {2} = \frac {y}{p _ {2}} - \frac {p _ {1}}{p _ {2}} x _ {1}$$

So $u(x_{1}) = \frac{y}{p_{2}} x_{1} - \frac{p_{1}}{p_{2}} x_{1}^{2}$.

Maximum of $u(x_{1})$ is attained at the point where $\frac{du}{dx_1} = 0 \rightarrow$

$$\begin{array}{l} \rightarrow \frac {y}{p _ {2}} - 2 \frac {p _ {1}}{p _ {2}} x _ {1} = 0 \rightarrow \\ \rightarrow x _ {1} = \frac {y}{2 p _ {1}}, \quad u _ {m a x} = u \left(\frac {y}{2 p _ {1}}\right) = \frac {y}{4 p _ {1} p _ {2}}. \end{array}$$

Demand functions:

$$x _ {1} = \frac {y}{2 p _ {1}}, \quad x _ {2} = \frac {y}{2 p _ {2}}.$$

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