Suppose that consumer's utility function for three goods is separable, that is,
$U(x_1, x_2, x_3) = f_1(x_1) + f_2(x_2) + f_3(x_3)$ ...(i)
where $f_i$ is increasing and strictly concave, i=1,2,3. Show that

  • $ df_i/dP_1 $ for i = 2,3 can be either positive or negative, but they must be of the same sign where $f_i$ is the demand function for good i and $P_1$ is price of good 1.

i reasoned using the equimarginal principle where at optimum we must have
$ MU_1/P1 = MU_2/P2 = MU_3/P3 $.
suppose good 1 is normal, if $P_1$ rises demand for good 1 falls, that is $df_1/dP_1$ < 0. Due to law of diminishing marginal utility $MU_1$ rises, however to maintain equilibrium $ MU_2, MU_3$ must also rise. So, $df_2/dp_1 < 0$ and $ df_3/dP_1 < 0$. But i couldn't do it more mathematically, taking the derivative of (i) w.r.t $P_1$ isn't of much use, please help.

  • $\begingroup$ You cannot "suppose good 1 is normal" or use "law of diminishing marginal utility". You are given a specific mathematical optimization problem (utility maximization w.r.t. budget constraint). Solve that, then reason from the solution. $\endgroup$
    – Giskard
    Commented Sep 9, 2018 at 15:07
  • 2
    $\begingroup$ Hint: Suppose $\mathrm df_2/\mathrm dp_1$ and $\mathrm df_3/\mathrm dp_1$ have different signs. See if you can deduce a contradiction, which would show that it's wrong to suppose the two derivatives have different signs. $\endgroup$
    – Herr K.
    Commented Sep 9, 2018 at 16:14
  • $\begingroup$ The question is odd. It is impossible for all derivatives to be positive. If one good becomes more expensive, you cannot suddenly buy more of all goods, at least if your income is independent of prices. $\endgroup$ Commented Sep 9, 2018 at 17:59
  • $\begingroup$ @denesp I can't suppose that good 1 is normal, but I can prove it (if a good is normal, when it's not inferior), since when consumer's utility function for goods are separable, it can be proved that none of the goods can be inferior. As for the use of law of diminishing marginal utility, it's given that $f_i$ is strictly concave, so I thought it was reasonable. I tried optimizing like you said, from the FOCs I arrived at the equimarginal principal, since the demand functions '$f_i(x_i)$ have not been explicitly mentioned, I couldn't proceed further. $\endgroup$
    – Chd
    Commented Sep 9, 2018 at 18:08
  • 1
    $\begingroup$ One more remark: The $f_i$ notation seems to be overladed, they are both demand functions and factors in the utility function. $\endgroup$ Commented Sep 9, 2018 at 18:54

1 Answer 1


The following argument assumes that we are dealing only with interior solutions. Let $(x_1,x_2,x_3)$ be an optimal demand bundle at prices $(p_1,p_2,p_3)$ and income $m$ and assume that $(x_1,x_2,x_3)$ is an optimal demand bundle at prices $(p_1',p_2,p_3)$ and income $m$. We want to show that one of the following three cases holds:

  1. $x_2=x_2'$ and $x_3=x_3'$,
  2. $x_2<x_2'$ and $x_3<x_3'$, or
  3. $x_2>x_2'$ and $x_3>x_3'$.

What is not allowed to happen is that $x_2\leq x_2'$ and $x_3>x_3'$ or $x_2>x_2'$ and $x_3\leq x_3'$. Assume one of these cases happens, without loss of generality, we can assume it is the first.

By the strict concavity of $f_2$ and $f_3$, we have $$MU_2(x_1,x_2,x_3)=\frac{\partial f_2(x_2)}{\partial x_2}\geq \frac{\partial f_2(x_2')}{\partial x_2}=MU_2(x_1',x_2',x_3')$$ and $$MU_3(x_1,x_2,x_3)=\frac{\partial f_3(x_3)}{\partial x_3}< \frac{\partial f_3(x_3')}{\partial x_3}=MU_3(x_1',x_2',x_3').$$ Therefore, $$\frac{MU_2(x_1',x_2',x_3')}{p_2}\leq\frac{MU_2(x_1,x_2,x_3)}{p_2}=\frac{MU_3(x_1,x_2,x_3)}{p_3}<\frac{MU_3(x_1',x_2',x_3')}{p_3},$$ in contradiction to the equimarginal principle.


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