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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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(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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    @xuan: I added a discussion for (b).2011-10-09