Question 11.9: The prices of inputs K and L are given as £12 per unit and £......

This question essentially asks us to demonstrate that the constrained maximization and minimization methods give consistent answers.
(i) The output constraint is that

1,250 = 25 K^{0.5}L^{0.5}

The objective function to be minimized is the cost function

TC = 12K + 3L

The corresponding Lagrange function is therefore

G = 12K + 3L + λ(1,250 − 25 K^{0.5}L^{0.5} )

First-order conditions require

\frac{\partial G}{\partial K } = 12 − λ12.5 K^{-0.5}L^{0.5} = 0      giving     \frac{12 K^{0.5}}{12.5L^{0.5}} = λ            (1)
\frac{\partial G}{\partial L} = 3 − λ12.5 K^{0.5}L^{-0.5} = 0      giving       \frac{3L^{0.5}}{12.5K^{0.5}} = λ            (2)
\frac{\partial G}{\partial \lambda } = 1,250 − 25 K^{0.5}L^{0.5} = 0                  (3)

Setting (1) equal to (2)

\frac{12 K^{0.5}}{12.5L^{0.5}}  = \frac{3L^{0.5}}{12.5K^{0.5}}
4K = L           (4)

Substituting (4) into (3)

1,250 − 25 K^{0.5}(4K)^{0.5} = 0
1,250 = 25 K^{0.5}(4)^{0.5}K^{0.5}
1,250 = 50K
25 = K

Substituting this value into (4)

4(25) = L
100 = L

When these optimum values of K and L are used the actual minimum cost will be

TC = 12K + 3L = 12(25) + 3(100) = 300 + 300 = £600

(ii) This part of the question requires us to find the values of K and L that will maximize output subject to a budget of £600, i.e. the answer to (i) above. The objective function to be maximized is therefore Q = 25 K^{0.5}L^{0.5} and the constraint is 12K + 3L = 600. The corresponding Lagrange equation is thus

G = 25 K^{0.5}L^{0.5} + λ(600 − 12K − 3L)

First-order conditions require

\frac{\partial G}{\partial K} = 12.5 K^{-0.5}L^{0.5} − 12λ = 0     giving     \frac{12.5L^{0.5}}{12K^{0.5}} = λ      (5)
\frac{\partial G}{\partial L} = 12.5 K^{0.5}L^{-0.5} − 3λ = 0      giving     \frac{12.5K^{0.5}}{3L^{0.5}} = λ        (6)
\frac{\partial G}{\partial \lambda }  = 600 − 12K − 3L = 0        (7)

Setting (5) equal to (6)

\frac{12.5L^{0.5}}{12K^{0.5}}  = \frac{12.5K^{0.5}}{3L^{0.5}}
3L = 12K
L = 4K         (8)
Substituting (8) into (7)

600 − 12K − 3(4K) = 0
600 − 12K − 12K = 0
600 = 24K
25 = K

Substituting this value into (8) L = 4(25) = 100
These are the same optimum values of K and L that were found in part (i) above. The actual output produced by 25 of K plus 100 of L will be

Q = 25 K^{0.5}L^{0.5} = 25 (25)^{0.5}(100)^{0.5} = 1,250

which checks out with the amount specified in the question.