Econ 310: Microeconomic Theory IISolutions to Problem Set 1

Solution 1 First, set up the Lagrangian:

L = x14y

14 + λ (20− 2x− 5y) ,

where λ is the Lagrangian multiplier. Second, take the first-order conditions:∂L∂x

=1

4x−

34y

14 − 2λ = 0 (1)

∂L∂y

=1

4x14y−

34 − 5λ = 0 (2)

∂L∂λ

= 20− 2x− 5y = 0. (3)

Dividing (1) by (2), we get14x−

34y

14

14x14y−

34

=2λ

5λ⇒ y

x=2

5.

Using the above to substitute out y in (3), we solve for x = 5. Hence, y = 2.

Solution 2 First, set up the Lagrangian:

L = x15y

15 z

15 + λ (30− x− 3y − 2z) ,

Second, take the first-order conditions:∂L∂x

=1

5x−

45y

15 z

15 − λ = 0 (4)

∂L∂y

=1

5x15y−

45 z

15 − 3λ = 0 (5)

∂L∂z

=1

5x15y

15 z−

45 − 2λ = 0 (6)

Third, impose the complementary slackness condition. (i) If λ = 0. Then(4) implies that either y = 0 or z = 0 or both, which contradicts y, z >0. Hence we can rule out this case. (ii) If λ > 0, which implies that theconstraint holds with equality (by complementary slackness), i.e.,

x+ 3y + 2z = 30. (7)

Dividing (4) by (5), we get x = 3y. Dividing (4) by (6), we get x = 2z.Substituting these into (7), we can solve for x = 10. Thus y = 10

3and z = 5.

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