# Economic optimisation problem

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? - You seem to be relatively new to MathSE, so I wanted to let you know a few things. Posting questions in the imperative (i.e. Compute all such, Prove that...), is considered rude by some of the members, so it would be nice of you to change that wording; perhaps by adding what are your thoughts or what you have tried in trying to answer the problem. If this is homework, please add the [homework] tag; don't worry, it won't stop people form answering. These sort of pleasantries usually result in more and better answers. Thank you. – Arturo Magidin Oct 9 '11 at 5:54 Thank you for your advice. I edited my question =) – xuan Oct 9 '11 at 6:29 @xuan: I observe that you have not accepted an answer to any of the questions you have asked. Please accept answers that satisfy you by clicking the tick. By the way, you can up- and downvote questions and answers by clicking the arrows on the left of the post. – Rasmus Oct 9 '11 at 15:56 ## 1 Answer (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