# 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? - ## 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