4
$\begingroup$

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?

|improve this question
$\endgroup$
5
$\begingroup$

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.

Update

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 $0$.

|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.