Demand Function

Question

A consumer has the utility function over goods X and Y, $U(X; Y) = X^{1/3}\cdot Y^{1/2}$

Let the price of good x be given by $P_x$, let the price of good $y$ be given by $P_y$, and let income be given by $I$.

(a) Derive the consumer’s generalized demand function for good $X$. This is simply the demand equation. $X$ is a function of Price and Income ($P_x$ and $I$).

Can someone help me get in the right direction for this?

Answer 1

You need to solve the following: $$\max_{X,Y}X^{1/3}Y^{1/2}$$ subject to $I \geq P_x X + P_y Y.$

Set up Lagrangian $$L = X^{1/3}Y^{1/2} + \lambda (I -(P_x X + P_y Y)).$$

FOC's yield: \begin{eqnarray*} \frac{1}{3}\left(\frac{Y^{1/2}}{X^{2/3}}\right) - \lambda P_x &=&0,\\ \frac{1}{2}\left(\frac{X^{1/3}}{Y^{1/2}}\right) - \lambda P_y &=&0,\\ I - (P_x X + P_y Y) &=&0. \end{eqnarray*} I will leave the algebra for you. You should get the following answer: $$X= \frac{2I}{5 P_x}.$$

Answer 2

You can solve the problem by using the method of lagrange multiplier as you have already written in the comment.

$$\mathcal L=X^{1/3}\cdot Y^{1/2}+\lambda (I-P_x\cdot X-P_y\cdot Y)$$

Then you have to calculate the partial derivatives w.r.t $X,Y$ and $\lambda$.

$\frac{\partial \mathcal L}{\partial X}=\frac13\cdot X^{-2/3}\cdot Y^{1/2}-P_X\lambda=0 \Rightarrow\frac13\cdot X^{-2/3}\cdot Y^{1/2}=P_x\lambda \quad (1)$

$\frac{\partial \mathcal L}{\partial Y}=\frac12\cdot X^{1/3}\cdot Y^{-1/2}-P_y\lambda=0 \Rightarrow\frac12\cdot X^{1/3}\cdot Y^{-1/2}=P_y\lambda \quad (2)$

$\frac{\partial \mathcal L}{\partial \lambda}=I-P_x\cdot X-P_y\cdot Y=0 \quad (3)$

Dividing (1) by (2)

$\frac23\frac{Y}{X}=\frac{P_x}{P_y}$

Solving for $P_y\cdot Y$

$P_y\cdot Y=\frac32P_x\cdot X$. The expression can be insert in (3).

$I-P_x\cdot X-\frac32P_x\cdot X=0$

What is left is to solve the equation for $X$. Can you finish?