Question

A consumer consumes two agricultural products: Red Meat, and Tomatoes according to the following utility function:
U = RT
Write down the budget constraint (budget line equation) for this consumer.

Determine the quantities that the consumer should consume of each of the two products.


Calculate the value of the lagrangian multiplier.
Derive the demand function for Red Meat.
Derive the demand function for Tomatoes.

Answer 1

Let y  be the income of the consumer and let p_{R ,} p_T be the prices of R and T. Then, given p_{R ,} p_T , if R and T  are the quantities consumed of the two goods, then, they should not cost more than y.

a)

So, the budget constraint is: _pRR +_pTT = y 

We are told that y = 210, p_R = 10 and p_T = 2

So the budget constraint becomes: $10R +2T = 210$

b)

Now we can move on to find the optimum quantities to be consumed. We can do that in two ways:

1) Using the fact that, at the optimum, The ratio of Marginal Utilities is equal to the price, and then using the budget constraint to solve for optimum quantities:

We want

$\frac{MU_R} {MU_ T} =\frac{p R} {p T} =\frac{10} {2} =5$

Also, recall that $MU_ x =\frac{∂U} {∂x}$

So, we have

$\frac{MU_R} {MU_ T} = \frac{\frac{∂U} {∂R}}{\frac{∂U} {∂T}}=\frac{\frac{∂(RT)}{ ∂R}} {\frac{∂(RT)} {∂T}} =\frac{T}{ R} = 5 \\T = 5R$

From the budget constraint we have,

$10R+2T=210$

Now, using the above equation with the budget constraint:

$10R+2T=210 \& T = 5R$

$⇒10R + 2T = 10R+2(5R)=20R=210$

$\\ ⇒20R = 210$

$\\ ⇒R = \frac{210}{20}=10.5$

$\\ ⇒T = 5R = 52.5$

2) We cam also maximize the give utility function subject to the Budget constraint using the Lagrangian multiplier

We will solve using variables instead of values and check if the solution is same as above. This will also give us the demand functions required in part c):

c) Setting up the Lagrangian:

$ℒ = RT − λ(_{pR}R+_{pT}T−y)$

Taking derivatives to find the first order conditions(FOCs):

$\frac{∂ℒ}{∂R}=T − λpR=0$

$\\ \frac{∂ℒ}{∂T}=R − λpT=0$

$\\ \frac{∂ℒ}{∂λ}=_{pR}R+_{pT}T−y=0$

Taking the ratio of the first two FOCs we have:

$\frac{T}{R}=\frac{pR}{pT}⇒_{pT}T=_{pR}R$

$\\⇒_{pR}R = y−_{pT}T$

From the third FOC, we have

$_{pR}R+_{pT}T−y=0$

Substituting this above,

$_{pT}T=_{pR}R \& _{pR}R = y−_{pT}T$

$\\⇒2_{pT}T=y$

$\\⇒T=\frac{y}{2_{pT}}$

and since

$_{pT}T=_{pR}R⇒R=\frac{pT}{pR}T$

$\\⇒R=\frac{pT}{pR}T =\frac{pT}{pR}\frac{y}{2_{pT}} =\frac{y}{2_{pR}}$

So the demand functions are:

$T = \frac{y}{2_{pT}} , R=\frac{y}{2_{pR}}$

 And the lagrange multiplier is, from the first FOC:

$λ=\frac{T}{pR}=\frac{y}{2_{pTpR}}$

Let us check if this demand function gives us same values as above:

$pT=2, pR=10, y=210$

$\\So,$

$\\ T = \frac{y}{2_{pT}}=\frac{210}{2×2}=\frac{210}{4}=52.5$

$\\R=\frac{y}{2_{pR}}=\frac{210}{2×10}=\frac{210}{20}=10.5$

$\\also,$

$\\ λ=\frac{y}{2_{pTpR}}=\frac{210}{2×2×10}=5.25$

We see that the demand functions give us same optimum values as above!