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Please look at only part ii.

I wrote budget constraint $x_1+x_2=10-f$

When I don’t pay fee, my optimal value values${}=(1,1).$

In the case of free disposal, $b_1=b_2=5$ but I consume only $x_1=0$ and $x_2=5.$

However I couldn’t find the value of $f.$ How can I find?

  • $\begingroup$ Hint: To be able to pay for free disposal you cannot use all income to buy goods. $\endgroup$ Jun 13, 2018 at 8:14
  • $\begingroup$ Hint 2: With free disposal you need to find the cheapest option that outperforms the optimal value you found when not paying the fee. $\endgroup$ Jun 13, 2018 at 8:16
  • $\begingroup$ Hint 3: The consumer will be willing to spend his remaining income on free disposal. $\endgroup$ Jun 13, 2018 at 8:17

1 Answer 1


Consumer's problem in default scenario:

\begin{eqnarray*} \max_{x_1, x_2} & \ \frac{x_2}{(1+x_1)^2} \\ \text{s.t.} & \ x_1 + x_2 \leq 10 \\& \ 0 \leq x_2 \leq x_1 \end{eqnarray*}

We can easily show that the solution will satisfy $x_2 = x_1$, and therefore we can rewrite the above problem as:

\begin{eqnarray*} \max_{x_1} & \ \frac{x_1}{(1+x_1)^2} \\ \text{s.t.} & \ 0 \leq x_1 \leq 5 \end{eqnarray*}

Solving this problem we get $x_1= 1$. Consequently, $x_2 = 1$ and the utility of the consumer in optimum is $\dfrac{1}{4}$.

Consumer's problem with a possibility of freely disposing good 1 after paying the fee:

\begin{eqnarray*} \max_{x_1, x_2} & \ x_2 \\ \text{s.t.} & \ x_1 + x_2 \leq m \\& \ 0 \leq x_2 \leq x_1 \end{eqnarray*} where $m$ is the net income of the consumer after paying the lump-sum fee.

Solution to this problem is $\left(\dfrac{m}{2}, \dfrac{m}{2}\right)$, and the utility of the consumer in optimum is $\dfrac{m}{2}$. Comparing this with the default scenario yields that consumer is indifferent between the scheme and the default when $m = \frac{1}{2}$. Therefore, the maximum willingness to pay to dispose the good at $0$ price is $10 - \frac{1}{2} = 9.5$.


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