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?
 A: (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
$$
p=\frac{1}{45}c^2.
$$
