When graduate exam questions require both computation and interpretation, it helps to learn the method as a chain of justified steps. Online tutoring can help you practice that reasoning under time constraints.
This article offers a graduate-level explanation centered on definitions, assumptions, and step-by-step logic. The goal is to clarify how the method works and how to apply it correctly, including where common misunderstandings occur. The discussion is designed to be accurate, self-contained, and coursework-ready.
For broader coverage, see Microeconomics Tutoring or the parent page Economics Tutoring. More worked examples are collected in the Economics Post Hub. If you’re stuck on Lagrangians, FOCs, or tangency, visit Troubleshooting: Theory.
Problem
A consumer maximizes CRRA utility:
$$ u(x,y) = \frac{x^{1-\gamma}}{1-\gamma} + \frac{y^{1-\gamma}}{1-\gamma}, \quad \gamma>0,\; \gamma\neq 1. $$
Let $\gamma=2$. Then:
$$ u(x,y)=-x^{-1}-y^{-1}. $$
Prices are $p_x=2$ and $p_y=1$. Income is $m=60$. The budget constraint is:
$$ 2x+y=60. $$
Graduate students may encounter this topic in coursework, exams, or research-related assignments. Reviewing both intuition and formal steps can improve understanding. Online tutoring support is available for graduate-level studies.
Solution
1) Lagrangian
$$ \mathcal{L} = -x^{-1} – y^{-1} + \lambda(60-2x-y). $$
2) First-order conditions
$$ \frac{\partial\mathcal{L}}{\partial x} = x^{-2}-2\lambda = 0 \quad\Rightarrow\quad \lambda=\frac{1}{2x^2}. $$
$$ \frac{\partial\mathcal{L}}{\partial y} = y^{-2}-\lambda = 0 \quad\Rightarrow\quad \lambda=y^{-2}. $$
$$ \frac{\partial\mathcal{L}}{\partial\lambda} = 60-2x-y = 0. $$
3) Solve
$$ \frac{1}{2x^2}=y^{-2} \quad\Rightarrow\quad y=\sqrt{2}\,x. $$
$$ x(2+\sqrt{2})=60 \quad\Rightarrow\quad x^*=\frac{60}{2+\sqrt{2}}, \qquad y^*=\sqrt{2}\,x^*. $$
4) Tangency check
$$ MRS = \left(\frac{y}{x}\right)^2 = 2 = \frac{p_x}{p_y}. $$
Conclusion:
$$ x^*=\frac{60}{2+\sqrt{2}}, \qquad y^*=\sqrt{2}\,x^*. $$
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