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

<p>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.</p>

<p><strong>Update</strong> </p>

<p>Let's add the <em>either a or b</em> and see what happens:</p>

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

<p>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 $0$.</p>

CC BY-SA 3.0
2015-02-11T12:06:04.457Z
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?
saubhik
CC BY-SA 3.0
2015-02-11T13:19:21.710Z
@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.
FooBar
CC BY-SA 3.0
2015-02-11T11:42:33.167Z
2015-02-11T13:18:21.157Z
2015-02-11T13:18:21.157Z
FooBar
<p>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 $40$.</p>

<p>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 $a=b=10$.</p>

<p>Am I right here?</p>

<microeconomics><consumer-theory>
Solving for optimal consumption bundle
saubhik
# Solving for optimal consumption bundle

(Asked by: username_0 on 2015-02-11)

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?

**Tags:** microeconomics, consumer-theory

---

## 1 answers

### Answer by username_1 on 2015-02-11. Score: 5 | Accepted answer

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 $0$.

> **Comments:**
> **username_0**: 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?  
> **username_1**: @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.  

---

