Answer to Question #242336 in Macroeconomics for mwesh

Question

Suppose the utility function for Osa is given U = f (Y, X) = X^0.5 Y^0.5. Given that the price of commodity X is Kshs. 10 per unit and that of community Y is Kshs. 12 per unit. Using Lagrangian multiplier technique determine units of the two commodities Osa will buy given an income of Kshs. 48,000 in order to maximize her utility, then compute both the values of U and lambda and provide an economic interpretation for lambda.

Answer 1

$Px= 10 \\ py = 12 \\ M = 48000 \\ Budget\ Constraint\\ XPx + YPy = M \\10X + 12Y = 48000$

$Maximize\ X\\ ^0\\^.\\^5 \ Y\\ ^0\\^.\\^5 \\ Subject\ to\ 10X + 12Y = 48000$

$Setitng \ up \ Lagrange:\\ L =X\\ ^0\\^.\\^5 \ Y\\ ^0\\^.\\^5+ \lambda(48000 - 10X - 12Y)$

$\frac{dL}{dX} =X\\ ^0\\^.\\^5 \ Y\\ ^0\\^.\\^5 - 10\lambda = 0 ........................(1)$

$\frac{dL}{dY} =X\\ ^0\\^.\\^5 \ Y\\ ^-\\^0\\^.\\^5 - 12\lambda = 0 ........................(2)$

$\frac{dL}{d\lambda} =48000 - 10X - 12Y = 0 ......................(3)$

Dividing (1) and (2):

$X\\ ^0\\^.\\^5 \ Y\\ ^-\\^0\\^.\\^5 / X\\ ^0\\^.\\^5 \ Y\\^0\\^.\\^5 = \frac{10\lambda}{12\lambda}$

$\frac{Y}{X} = \frac{5}{6}$

$Y = \frac{5X}{6}$

Substituting the value of Y in (3)

$48000 - 10X - 12\times \frac{5X}{6} = 0$

$48000 - 20X = 0$

$X = 2400$

$Y = 5 \times \frac{2400}{6}$

$Y = 2000$

$U = X\\ ^0\\^.\\^5 \ Y\\ ^0\\^.\\^5$

$U = 2400\\ ^0\\^.\\^5 \ 2000\\ ^0\\^.\\^5$

$U = 2191$

From (1)

$0.5Y\\ ^0\\^.\\^5 \ Y\\ ^0\\^.\\^5 - 10\lambda = 0$

$0.5 \times 2400\\ ^0\\^.\\^5 \ 2000\\ ^0\\^.\\^5 - 10\lambda = 0$

$10\lambda = 0.456$

$\lambda = 0.0456$

Economic interpretation:

$At\ the\ utility\ maximizing\ point\ the\ marginal\ utility\ per\ dollar\\ obtained\ from\ the\ consumption\ of\ X\ equal\ the\ marginal\ utility\\ \ per\ dollar\ obtained\ from\\ the\ consumption\ of\ Y\ which\ is\ equal\ to\ \lambda, i.e., 0.0456$