You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
messy algebra from optimization problem
- Thread starter woolley
- Start date
The original problem:
What condition(s) are sufficient to ensure that the second order conditions are satisfied for the
optimization problem below? You can express this condition in terms of \(\displaystyle \alpha, \beta, \gamma, \delta\) and the
values of x and y that satisfy the first order conditions that solve the problem below, x* and y*,
but not in terms of \(\displaystyle \lambda^*\). There is no need to evaluate determinants – state condition in terms of
determinants if you desire.
\(\displaystyle \max_{x,y} \;x^\alpha+y^\beta \qquad s.t.\;\; x^\gamma+y^\delta<b\)
What i have done:
\(\displaystyle $f(x,y)=x^\alpha+y^\beta$ and $g(x,y)=x^\gamma+y^\delta$ with $0<\alpha,\beta,\gamma,\delta<1$\[\mathcal{L}=x^\alpha+y^\beta+\lambda(x^\gamma+y^\delta-b)\]FOC:\begin{align}\mathcal{L}_x: \alpha x^{\alpha-1}+\gamma \lambda x^{\gamma-1}=0\\\mathcal{L}_y: \beta y^{\beta-1}+\delta \lambda y^{\delta-1}=0\\\mathcal{L}_\lambda: x^\gamma+y^\delta-b=0\end{align} from (1) and (2) get \[x=\frac{\alpha\delta}{\gamma\beta}y^{\frac{\delta-\beta}{\gamma-\alpha}}\] substitute in (3) for \[b=\left(\frac{\alpha\delta}{\gamma\beta}y^{\frac{\delta-\beta}{\gamma-\alpha}}\right)^\gamma+y^\delta\] How do i solve for y in terms of \alpha, \beta, \gamma, \delta, \mathrm{and}\, b?\)
What condition(s) are sufficient to ensure that the second order conditions are satisfied for the
optimization problem below? You can express this condition in terms of \(\displaystyle \alpha, \beta, \gamma, \delta\) and the
values of x and y that satisfy the first order conditions that solve the problem below, x* and y*,
but not in terms of \(\displaystyle \lambda^*\). There is no need to evaluate determinants – state condition in terms of
determinants if you desire.
\(\displaystyle \max_{x,y} \;x^\alpha+y^\beta \qquad s.t.\;\; x^\gamma+y^\delta<b\)
What i have done:
\(\displaystyle $f(x,y)=x^\alpha+y^\beta$ and $g(x,y)=x^\gamma+y^\delta$ with $0<\alpha,\beta,\gamma,\delta<1$\[\mathcal{L}=x^\alpha+y^\beta+\lambda(x^\gamma+y^\delta-b)\]FOC:\begin{align}\mathcal{L}_x: \alpha x^{\alpha-1}+\gamma \lambda x^{\gamma-1}=0\\\mathcal{L}_y: \beta y^{\beta-1}+\delta \lambda y^{\delta-1}=0\\\mathcal{L}_\lambda: x^\gamma+y^\delta-b=0\end{align} from (1) and (2) get \[x=\frac{\alpha\delta}{\gamma\beta}y^{\frac{\delta-\beta}{\gamma-\alpha}}\] substitute in (3) for \[b=\left(\frac{\alpha\delta}{\gamma\beta}y^{\frac{\delta-\beta}{\gamma-\alpha}}\right)^\gamma+y^\delta\] How do i solve for y in terms of \alpha, \beta, \gamma, \delta, \mathrm{and}\, b?\)
Your equation labelled (3) and the equation that you obtained from (1) and (2) are conditions involving \(\displaystyle \alpha\), \(\displaystyle \beta\), \(\displaystyle \gamma\), \(\displaystyle \delta\), x* and y* (replace x and y), but not \(\displaystyle \lambda\). Isn't that all that is asked of you?woolley said:The original problem:
What condition(s) are sufficient to ensure that the second order conditions are satisfied for the
optimization problem below? You can express this condition in terms of \(\displaystyle \alpha, \beta, \gamma, \delta\) and the
values of x and y that satisfy the first order conditions that solve the problem below, x* and y*,
but not in terms of \(\displaystyle \lambda^*\). There is no need to evaluate determinants – state condition in terms of
determinants if you desire.