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Here is the question:

Consider a car-owning consumer with utility function $$u (x) = x_1x_2 + x_3 (x_4)^2 ,$$

where $x_1$ denotes food consumed, $x_2$ denotes alcohol consumed, $x_3$ denotes kms of city driving, and $x_4$ denotes kms of open road driving. Total fuel consumption (in litres) is given by $$h (x) = 2x_3 + x_4.$$

Food costs \$1 per unit, alcohol costs \$1 per unit and the consumer has income of \$10.

(a) Suppose fuel is provided free by the Government, but is rationed. This consumer is allowed to use c litres of fuel. Find her optimal choice of $x_1$, $x_2$, $x_3$ and $x_4$.

(b) Now suppose the consumer receives her quota of fuel ($c$ litres), but can buy or sell fuel on a fuel market for $p$ per litre. Find the value of $p$ at which this consumer will choose not to trade on the fuel market.

I tried writting the Lagrangian $L(x_1,x_2,x_3,x_4,\lambda,\mu) = x_1x_2 + x_3(x_4)^2 - \lambda(2x_3+x_4-c) - \mu(x_1+x_2-10)$. Is this correct? If so, how do I solve this equation? When I set the derivatives to $0$, I get $x_1=x_2=\mu=5$ but I can't solve for $x_3$, $x_4$ or $\lambda$.(resolved)

Now I have solved part (a), but how to solve part (b)? Is the price just equal to λ from part (a) or do I have to write a new Lagrangian?

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up vote 1 down vote accepted

(a) You corrently determined the optimal values for $x_1$ and $x_2$. I will show you how to do $x_3$ and $x_4$.

Setting the partial derivatives with respect to $x_3$ and $x_4$ equal to zero we get $$ x_4^2-2\lambda=0, $$ $$ 2x_3 x_4-\lambda=0. $$ Combining these we get $$x_4=4x_3.$$

And we still have the constraint $$ 2x_3+x_4=c. $$

Solving this system of linear equations we get $$ x_3=\frac{1}{6}c, $$ $$ x_4=\frac{2}{3}c. $$ The value of $\lambda$ is then $\frac{2}{9}c^2$, and the optimized utility is $25+\frac{2}{27}c^3$.

(b) Assuming that the comsumer buys $b$ litres of fuel, we have to replace $c$ with $c+b$ and $10$ with $10-pb$ in the computation we did in (a).

Proceeding exactly as in (a), we get the following optimal values for the $x_i$: $$ x_1=x_2=\frac{10-pb}{2}, $$ $$ x_3=\frac{1}{6}(c+b), $$ $$ x_4=\frac{2}{3}(c+b). $$ The optimized (with respect to the $x_i$) utility is then $$ x_1 x_2+x_3 x_4^2=\Bigl(\frac{10-pb}{2}\Bigr)^2+\frac{2}{27}(c+b)^3. $$

If it is optimal to not buy or sell any full, then the derivarive with respect to $b$ of the above expression is $0$. The derivarive is $$ -p(10-pb)+\frac{2}{9}(c+b)^2. $$ Setting $b=0$ gives $$ \frac{2}{9}c^2-10p. $$ Hence our consumer will not want to buy or sell fuel if $$ p=\frac{1}{45}c^2. $$

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Thank you Rasmus! It was really helpful. Do you know how to solve part b)? – xuan Oct 9 '11 at 9:14
@xuan: I added a discussion for (b). – Rasmus Oct 9 '11 at 10:01

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