Question

Suppose that the manager of a firm is planning to meet an order of 1000 units of two products X and Y. The manager's problem is to find the combination of two goods that minimize its cost. He has the firm's cost function of two goods estimated as

C = 5X2 + 20 Y2

By using the Lagrangian multiplier method, find the quantity of X and quantity of Y, subject to X + Y = 1000, that minimize the cost of meeting the order.

Answer 1

$C=5X^2+20Y^2$

X+Y=1000

Langrangian

$L= C+\lambda(1000-X-Y)$

$=$ $5X^2+20Y^2+\lambda(1000-X-Y)$

First order Condition

$\frac{\Delta L}{\Delta X}=10X-\lambda=0$ ..........(i)

$X=\frac{\lambda}{10}$

$\frac{\Delta L}{\Delta Y}=40Y-\lambda=0$ ...........(ii)

$Y=\frac{\lambda}{40}$

$\frac{\Delta L}{\Delta\lambda}=1000-X-Y=0$ ........(iii)

X+Y=1000

Using Eqns (i)) and (ii)

$\frac{\lambda}{10}+\frac{\lambda}{40}=1000$

$\frac{(4\lambda+\lambda)}{40}=1000$

$\lambda=\frac{40,000}{5}= 8000$

$X=\frac{8000}{10}= 800$

$Y=\frac{8000}{40}= 200$