Suppose that Maria cares only about shoes and clothes. Her utility function is U = S^0.5C^0.5, where S is the number of pairs of shoes and C is the number of clothing items that she owns. The price of shoes is $80, and the price of clothes is $100. Suppose that Maria must have 1,000 units of utility and wants to achieve this level of utility with the lowest possible expenditure. How can Maria's expenditure minimization problem be expressed as a Lagrangian equation?

Question

Quizplus · Accepted Answer

To express Maria's expenditure minimization problem as a Lagrangian equation, we need to set up the following optimization problem:

Minimize E = 80S + 100C

Subject to:

(U - SS1U1P10.5S1S1P0CS1U1P10.5S1S1P0) = 0

To write this in the Lagrangian form, we need to introduce a Lagrange multiplier λ and rewrite the above constraint as:

f (S, C, λ) = 80S + 100C + λ(1000 - SS1U1P10.5S1S1P0CS1U1P10.5S1S1P0)

Now, we need to find the values of S, C, and λ that minimize the above function f. This can be done by taking the partial derivatives of f with respect to S, C, and λ, and setting them equal to 0:

∂f / ∂S = 80 - λ(0.5U1P1-0.5S1-0.5) = 0

∂f / ∂C = 100 - λ(0.5U1P1-0.5S1-0.5) = 0

∂f / ∂λ = 1000 - SS1U1P10.5S1S1P0CS1U1P10.5S1S1P0 = 0

Solving these equations simultaneously, we get:

S = (U1P1-0.5S1-0.5) / 2

C = (U1P1-0.5S1-0.5) / 2

λ = (U1P10.5S1S1P0CS1U1P10.5S1S1P0) / 1000

Substituting these values back into the original equation for f, we get:

f = (U1P1-0.5S1-0.5) / 2 x 80 + (U1P1-0.5S1-0.5) / 2 x 100 + [(U1P10.5S1S1P0CS1U1P10.5S1S1P0) / 1000] x (1000 - U)

Simplifying this expression, we get:

f = 40U1P1-20S1-20 + 50U1P1-25S1-25 + 0.1(U1P10.5S1S1P0CS1U1P10.5S1S1P0)(1000 - U)

Therefore, the Lagrangian equation for Maria's expenditure minimization problem is:

f = 40U1P1-20S1-20 + 50U1P1-25S1-25 + 0.1(U1P10.5S1S1P0CS1U1P10.5S1S1P0)(1000 - U)

which corresponds to answer choice C.

Suppose That Maria Cares Only About Shoes and Clothes