<p>If we are going to consider the separate effects on consumer surplus (CS) of <span class="math-container">$q_1$</span> and <span class="math-container">$q_2$</span>, what we must do is first find the CS when <span class="math-container">$Q=q_1$</span> and then consider the <em>total change</em> <em>in CS</em> which results from increasing <span class="math-container">$Q$</span> from <span class="math-container">$q_1$</span> to <span class="math-container">$q_1+q_2$</span>. In the diagram below, the CS when <span class="math-container">$Q=q_1$</span> is represented by area <span class="math-container">$A$</span> which equals <span class="math-container">$(q_1/2)^2$</span>.</p>
<p><a href="https://i.sstatic.net/QD5R2.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/QD5R2.png" alt="enter image description here" /></a></p>
<p>The increase from <span class="math-container">$q_1$</span> to <span class="math-container">$q_1+q_2$</span> has two effects on CS. It adds area C which equals <span class="math-container">$q_2/2)^2$</span>. But also, by lowering the price, including the price for the original <span class="math-container">$q_1$</span> units, it increases the CS on those units.  This second effect is represented by area B, a rectangle with length <span class="math-container">$q_1$</span> and height <span class="math-container">$q_2/2$</span> and therefore area <span class="math-container">$q_1q_2/2$</span>.  Thus the total CS obtained in this way is as below:</p>
<p><span class="math-container">$$CS = \Big(\frac{q_1}{2}\Big)^2+\frac{q_1q_2}{2}+\Big(\frac{q_2}{2}\Big)^2 =\frac{q_1^2+2q_1q_2+q_2^2}{4} = \Big(\frac{q_1+q_2}{2}\Big)^2$$</span></p>
<p>which equals <span class="math-container">$(Q/2)^2$</span> when <span class="math-container">$Q=q_1+q_2$</span>.</p>
<p>So the answer to your question &quot;Why's this happening?&quot; is that your calculation ignores area B, the effect of producing additional units on the consumer surplus of existing units.</p>

CC BY-SA 4.0
2024-04-06T21:48:00.073Z
That works providing you assume machine 1 comes first providing consumer surplus A and then machine 2 comes along to add C and B.  Arguably each machine provides B in addition to A or C : if machine 1 was not there then machine 2 would provide C, so machine 1 increases consumer surplus by A and B.  Then the problem becomes trying to avoid overcounting B,
Henry
CC BY-SA 4.0
2024-04-06T22:25:49.857Z
@Henry  Yes, in the interests of brevity I only considered machine 1 coming first, but the argument as you indicate is easily adapted for machine 2 coming first.
Adam Bailey
CC BY-SA 4.0
2024-04-07T04:42:33.320Z
@Henry Right, and if both the machines work together, then we simply think of it as the final quantity $Q$ (rather than separating them).
consumer-surplus
CC BY-SA 4.0
2024-04-06T19:23:58.680Z
2024-04-06T19:37:48.143Z
2024-04-06T19:37:48.143Z
Adam Bailey
<p>Suppose <span class="math-container">$P(Q) = 10-0.5Q$</span>. If a firm is producing <span class="math-container">$q_1$</span> and <span class="math-container">$q_2$</span> units from its two different machines such that <span class="math-container">$Q = q_1 + q_2$</span>, what is the consumer surplus at <span class="math-container">$(q_1, q_2)$</span>?</p>
<p>I think the consumer surplus is <span class="math-container">$\left(\frac{Q}{2}\right)^2$</span>.</p>
<p>However, when I calculate the consumer surpluses separately, the result is different. In other words, at <span class="math-container">$q_i$</span>, the consumer surplus is <span class="math-container">$(q_i/2)^2$</span> but the total consumer surplus <span class="math-container">$(q_1/2)^2 + (q_2/2)^2 \neq (Q/2)^2$</span>.</p>
<p>Why's this happenning?</p>

CC BY-SA 4.0
2024-04-06T16:06:07.897Z
Why would you "calculate the surpluses separately" for each production machine and then add them together? They are related to how demand varies with price. You might calculate  consumer surpluses separately for each consumer, but not for each producer.
Henry
CC BY-SA 4.0
2024-04-06T17:13:21.330Z
@Henry Why can't we do that? I can see it does not work but I don't see why.
consumer-surplus
<microeconomics><consumer-theory><consumer-surplus>
Consumer surplus calculation
consumer-surplus
# Consumer surplus calculation

(Asked by: username_0 on 2024-04-06)

Suppose $P(Q) = 10-0.5Q$. If a firm is producing $q\_1$ and $q\_2$ units from its two different machines such that $Q = q\_1 + q\_2$, what is the username_0 at $(q\_1, q\_2)$?

I think the username_0 is $\left(\frac{Q}{2}\right)^2$.

However, when I calculate the consumer surpluses separately, the result is different. In other words, at $q\_i$, the username_0 is $(q\_i/2)^2$ but the total username_0 $(q\_1/2)^2 + (q\_2/2)^2 \neq (Q/2)^2$.

Why's this happenning?

**Tags:** microeconomics, consumer-theory, consumer-surplus

> **Comments:**
> **username_1**: Why would you "calculate the surpluses separately" for each production machine and then add them together? They are related to how demand varies with price. You might calculate  consumer surpluses separately for each consumer, but not for each producer.  
> **username_0**: @username_1 Why can't we do that? I can see it does not work but I don't see why.  

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## 1 answers

### Answer by username_2 on 2024-04-06. Score: 1 | Accepted answer

If we are going to consider the separate effects on username_0 (CS) of $q\_1$ and $q\_2$, what we must do is first find the CS when $Q=q\_1$ and then consider the *total change* *in CS* which results from increasing $Q$ from $q\_1$ to $q\_1+q\_2$. In the diagram below, the CS when $Q=q\_1$ is represented by area $A$ which equals $(q\_1/2)^2$.

[![enter image description here](https://i.sstatic.net/QD5R2.png)](https://i.sstatic.net/QD5R2.png)

The increase from $q\_1$ to $q\_1+q\_2$ has two effects on CS. It adds area C which equals $q\_2/2)^2$. But also, by lowering the price, including the price for the original $q\_1$ units, it increases the CS on those units. This second effect is represented by area B, a rectangle with length $q\_1$ and height $q\_2/2$ and therefore area $q\_1q\_2/2$. Thus the total CS obtained in this way is as below:

$$CS = \Big(\frac{q\_1}{2}\Big)^2+\frac{q\_1q\_2}{2}+\Big(\frac{q\_2}{2}\Big)^2 =\frac{q\_1^2+2q\_1q\_2+q\_2^2}{4} = \Big(\frac{q\_1+q\_2}{2}\Big)^2$$

which equals $(Q/2)^2$ when $Q=q\_1+q\_2$.

So the answer to your question "Why's this happening?" is that your calculation ignores area B, the effect of producing additional units on the username_0 of existing units.

> **Comments:**
> **username_1**: That works providing you assume machine 1 comes first providing username_0 A and then machine 2 comes along to add C and B.  Arguably each machine provides B in addition to A or C : if machine 1 was not there then machine 2 would provide C, so machine 1 increases username_0 by A and B.  Then the problem becomes trying to avoid overcounting B,  
> **username_2**: @username_1  Yes, in the interests of brevity I only considered machine 1 coming first, but the argument as you indicate is easily adapted for machine 2 coming first.  
> **username_0**: @username_1 Right, and if both the machines work together, then we simply think of it as the final quantity $Q$ (rather than separating them).  

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