L3 - Optimal Choice II

Question 1

How to use the Lanrangian method to maximise someones utility?

Answer 1

1. Find state the Utility function and budget constraint
2. Then state as a Langrangian maximisation problem:

• L(q₁,q₂,λ) = U(q₁,q₂) + λ(Y -p₁q₁ - p₂q₂)
• OR
• L(q₁,q₂,λ) = U(q₁,q₂) - λ(p₁q₁ + p₂q₂-Y) (normally use this one)
• L=(Utility) + λ(Budget constraint equal to zero)

3. Find 3 first order condition or q₁,q₂ and λ:
   * dL/d(q₁),dL/d(q₂) and dL/dλ = 0
4. Solve simultanously till you get the optimal ratio between the two goods
5. insert Optimal ratio into Budget Contraint to solve for the quantities

Question 2

How does the Langarian Method line to the Interior solution?

Answer 2

L(q₁,q₂,λ) = U(q₁,q₂) - λ(p₁q₁ + p₂q₂-Y)

Taking the FOCs:

• dL/d(q₁) = dU/d(q₁) - λp₁ = 0
• dL/d(q₂) = dU/d(q2) - λp₂ = 0
• dL/dλ = -(p₁q₁ + p₂q₂-Y) = 0

Then the first two FOC imply (dU/d(q₁))/(dU/d(q2)) = λp₁/λp₂

Hence, this is where the condition MRS=MRT is derived from

Question 3

What is the interpretation of λ*?

Answer 3

Solving the FOC for λ* yields:

• (dU/d(q₁))/p_{1 =}(dU/d(q₂))/p₂
• When evaluated at q₁* and q₂*, this can give a number and can be interpreted as the marginal increase in utility that the consumer would gain if his income was increased by one unit – the marginal utility of income (marginal utility per pound)
• Why? If income went up by 1, the consumer could afford an extra (1/p₁) units of good X which would generate an increase in utility equal to MU_q1= dU / d(q₁) . Identical and equal argument for good Y. No way of getting this insight from graphical analysis!

Question 4

How does Second Order conditions apply when maximising utility?

Answer 4

• As always, when maximising a function, we have to be sure whether the stationary point we have found is a maximum rather than a minimum.
• With Lagrangians, this is a quite tedious and involves bordered Hessians etc. So don’t worry about all that.
• Fortunately, we also know that we have a maximum as long as the BC is linear and the IC’s are convex.
• If the IC was concave to the origin l, then we’d have a minimum.