Let $\min\{x,z\}=\Omega$, where $P_\Omega=P_x+P_z$. Now the problem becomes $U(x,\Omega)=y\Omega$, which is a standard Cobb-Douglas with degree 2 of homogeneity. Now in this case the choice for each good is:

$y^*=\frac{\alpha_y100}{P_y(\alpha_y+\alpha_\Omega)}\implies y^*=\frac{100}{2P_y}\;\;\;\;\;\;\;$in this case $\alpha_y=\alpha_\Omega=1$

For $\Omega$: $\;\;\;\;\;\Omega^*=\frac{\alpha_\Omega100}{P_\Omega(\alpha_y+\alpha_\Omega)}\implies \Omega^*=\frac{100}{2P_\Omega} \implies \Omega^*=\frac{100}{2(P_x+P_z)}$

Now, since $P_x=P_y=P_z$, let $P_x=P_y=P_z=P$ a general price, therefore substituting in our optimums:

$y^*=\frac{100}{2P}\;\;\;\;\;\Omega^*=\frac{100}{4P}$

Now it's straightforward (since we already know $x^*=z^*$ and as this is the optimum for $\min\{x,z\}$ which is $x$ OR $z$) that $y^*=\frac{100}{2P}>\Omega^*=\frac{100}{4P}$, so this implies that:

$y^*>x^*=z^*\;\;$****

Also I found this document, where this question is number 13.

Hope this helps.

Disclaimer: It would be helpful if other people could assess this approximation, since I hadn't never seen this problem before.

Let $\min\{x,z\}=\Omega$, where $P_\Omega=P_x+P_z$. Now the problem becomes $U(y,\Omega)=y\Omega$, which is a standard Cobb-Douglas with degree 2 of homogeneity. Now in this case the choice for each good is:

$y^*=\frac{\alpha_y100}{P_y(\alpha_y+\alpha_\Omega)}\implies y^*=\frac{100}{2P_y}\;\;\;\;\;\;\;$in this case $\alpha_y=\alpha_\Omega=1$

For $\Omega$: $\;\;\;\;\;\Omega^*=\frac{\alpha_\Omega100}{P_\Omega(\alpha_y+\alpha_\Omega)}\implies \Omega^*=\frac{100}{2P_\Omega} \implies \Omega^*=\frac{100}{2(P_x+P_z)}$

Now, since $P_x=P_y=P_z$, let $P_x=P_y=P_z=P$ a general price, therefore substituting in our optimums:

$y^*=\frac{100}{2P}\;\;\;\;\;\Omega^*=\frac{100}{4P}$

Now it's straightforward (since we already know $x^*=z^*$ and as this is the optimum for $\min\{x,z\}$ which is $x$ OR $z$) that $y^*=\frac{100}{2P}>\Omega^*=\frac{100}{4P}$, so this implies that:

$y^*>x^*=z^*\;\;$****

Also I found this document, where this question is number 13.

Hope this helps.

Disclaimer: It would be helpful if other people could assess this approximation, since I hadn't never seen this problem before.

Let $\min\{x,y\}=\Omega$, where $P_\Omega=P_x+P_y$. Now the problem becomes $U(x,\Omega)=x\Omega$, which is a standard Cobb-Douglas with degree 2 of homogeneity. Now in this case the choice for each good is:

$x^*=\frac{\alpha_x100}{P_x(\alpha_x+\alpha_\Omega)}\implies x^*=\frac{100}{2P_x}\;\;\;\;\;\;\;$in this case $\alpha_x=\alpha_\Omega=1$

For $\Omega$: $\;\;\;\;\;\Omega^*=\frac{\alpha_\Omega100}{P_\Omega(\alpha_x+\alpha_\Omega)}\implies \Omega^*=\frac{100}{2P_\Omega} \implies \Omega^*=\frac{100}{2(P_y+P_z)}$

Now, since $P_x=P_y=P_z$, let $P_x=P_y=P_z=P$ a general price, therefore substituting in our optimums:

$x^*=\frac{100}{2P}\;\;\;\;\;\Omega^*=\frac{100}{4P}$

Now it's straightforward (since we already know $y^*=z^*$ and as this is the optimum for $\min\{y,z\}$ which is $x$ OR $y$) that $x^*=\frac{100}{2P}>\Omega^*=\frac{100}{4P}$, so this implies that:

$x^*>y^*=z^*$

Also I found this document, where this question is number 13.

Hope this helps.

Disclaimer: It would be helpful if other people could assess this approximation, since I hadn't never seen this problem before.

Let $\min\{x,z\}=\Omega$, where $P_\Omega=P_x+P_z$. Now the problem becomes $U(x,\Omega)=y\Omega$, which is a standard Cobb-Douglas with degree 2 of homogeneity. Now in this case the choice for each good is:

$y^*=\frac{\alpha_y100}{P_y(\alpha_y+\alpha_\Omega)}\implies y^*=\frac{100}{2P_y}\;\;\;\;\;\;\;$in this case $\alpha_y=\alpha_\Omega=1$

For $\Omega$: $\;\;\;\;\;\Omega^*=\frac{\alpha_\Omega100}{P_\Omega(\alpha_y+\alpha_\Omega)}\implies \Omega^*=\frac{100}{2P_\Omega} \implies \Omega^*=\frac{100}{2(P_x+P_z)}$

Now, since $P_x=P_y=P_z$, let $P_x=P_y=P_z=P$ a general price, therefore substituting in our optimums:

$y^*=\frac{100}{2P}\;\;\;\;\;\Omega^*=\frac{100}{4P}$

Now it's straightforward (since we already know $x^*=z^*$ and as this is the optimum for $\min\{x,z\}$ which is $x$ OR $z$) that $y^*=\frac{100}{2P}>\Omega^*=\frac{100}{4P}$, so this implies that:

$y^*>x^*=z^*\;\;$****

Also I found this document, where this question is number 13.

Hope this helps.

Disclaimer: It would be helpful if other people could assess this approximation, since I hadn't never seen this problem before.