Consider a consumer who can consume either A or B, with the quantities being denoted by $a$ and $b$ respectively. If the utility function of the consumer is given by $$-[(10-a)^2+(10-b)^2]$$(suppose prices of both goods are equal to $1$), then solve for optimal consumption of the consumer when his income is $40$.

My approach: I have the problem: $$max(-[(10-a)^2+(10-b)^2])$$ $$s.t.\ a+b \le 40,\ a\ge 0,\ b\ge 0.$$ Looking at the objective function, we see that it's maximum value is $0$ when $a=b=10$.

Am I right here?


Yes, you are correct. That solution implies a utility of $0$, while any other solution necessarily will give you negative utility.

It is an odd problem for its violation of local nonsatiation: It is indeed optimal for the household to throw away the rest of his income.


Let's add the either a or b and see what happens:

$$max(-[(10-a)^2+(10-b)^2])$$ $$s.t.\ a+b \le 40,\ a\ge 0,\ b\ge 0,\ ab=0.$$

The optimal solution set now contains $\{(10, 0), (0, 10)\}$. The preferences between this one are still globally satiated at $(10,10)$, but as the point is not feasible, we set one of the coordinates to that value and keep the other one at $0$.

  • $\begingroup$ The phrase "either A or B" is making me doubtful. Does the consumer always consume exactly one of the two goods? Or am I thinking wrong? $\endgroup$
    – saubhik
    Feb 11 '15 at 12:06
  • 1
    $\begingroup$ @ponderlust right that slipped my attention. If we add it, the solution changes. I emphasize that $(10,10)$ is still the optimal solution. Him consuming exactly one of them is not what he would prefer, but apparently an additional restriction. $\endgroup$
    – FooBar
    Feb 11 '15 at 13:19

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