Help with CES Utility Function exercise

Question

I am attempting to solve the following CES Utility Function problem:

However, I am running into issues when I get to 3).

For 1) I have $K = \left(\frac{\omega p_1}{p_2}\right)^{\frac{1}{p+1}}$

For 2) I get $X_2^M = \frac{m}{\frac{p_1}{K}+p_2} $

For 3) I find $\lambda^* = (K^\rho + \omega)^{-\frac1p-1} \cdot \omega \cdot p_2^{-1} $

and $v(p_1, p_2, m) = \left(\left(\frac{m}{p_1+Kp_2}\right)^{-\rho} + \omega(\frac{mK}{p_1+Kp_2})^{-\rho}\right)^{-1/\rho}$

I then divide $\lambda^*$ by $v(p_1, p_2, m)$, but when I do so I can't seem to fully cancel out $p_1,p_2,m,K$ and $\rho$ which I believe I would have to do to prove that they are proportional. I'm not sure if the issue is with my $\lambda^*$, my $v(p_1,p_2,m)$ or both...

Additionally, for 6) how does one demonstrate homogeneity of a given degree?

Answer 1

For question 3, I'm sure there are simpler solution than mine. I provide this version for your reference.

By definition, \begin{align} \lambda^* &= \frac{\frac{\partial u}{\partial x_1}(x_1^M)}{p_1} = \frac{\left[(x_1^M)^{-\rho}+ w(x_2^M)^{-\rho}\right]^{-\frac{1}{\rho}-1}(x_1^M)^{-(\rho+1)}}{p_1}\\ \\ v(p_1,p_2,m) &= u(x_1^M(p_1,p_2, m), x_2^M(p_1, p_2, m)) \\ & = \left[(x_1^M)^{-\rho}+ w(x_2^M)^{-\rho}\right]^{-\frac{1}{\rho}} \end{align} So \begin{align} \lambda^* &= v(p_1,p_2,m) \frac{1}{p_1\left[(x_1^M)^{-\rho}+ w(x_2^M)^{-\rho}\right](x_1^M)^{\rho+1}}\\ &= v(p_1,p_2,m) \frac{1}{p_1\left[x_1^M+ w\left(\frac{x_1^M}{x_2^M}\right)^{\rho+1}x_2^M\right]}\\ & = v(p_1,p_2,m) \frac{1}{p_1\left[x_1^M+ w\left(\frac{1}{\kappa}\right)^{\rho+1}x_2^M\right]} \end{align} Plug in the expression for $\kappa = \left(\frac{wp_1}{p_2}\right)^{\frac{1}{\rho+1}}$ to the equation above, we have

\begin{align} p_1\left[x_1^M+ w\left(\frac{1}{\kappa}\right)^{\rho+1}x_2^M\right] &= p_1\left[x_1^M+ w\frac{p_2}{wp_1}x_2^M\right]\\ & = p_1x_1^M + p_2x_2^M\\ & = m \end{align} so $$\lambda^* = \frac{v(p_1,p_2,m)}{m}$$

To show a function $F(x_1, x_2, ... x_n)$ is homogenous of degree $k$, simply show that $$F(kx_1, kx_2, ...kx_n) = \lambda^kF(x_1, x_2, ... x_n)$$ For $\forall \;\lambda>0$