Finding Elasticity in Demand: Solving for p in the Demand Equation

[polak333]

Homework Statement

Derivatives and elasticity:
The demand equation for a product is q = \left(\frac{20-p}{2}\right)^{2} for 0 \leq p \leq 20.

a) find all values of p for which demand is elastic.

Homework Equations

Elasticity: \eta = \frac{p}{q} x \frac{dq}{dp}

The Attempt at a Solution

Well, I know that if it has to be elastic, then |η| > 1.
However, how do I show this.
Do I just set any number larger than 1 as η, and then solve for p and q?

 

[Mentallic]

Can you first show me what \eta is equal to?

 

[polak333]

Do you mean η = p / q · dq / dp

q = ((20-p)/2)^2
q' = (p-20)/2

.: η = p / ((20-p)/2)^2 · (p-20)/2

And since I know |η| must be greater than 1, then do I set η to something such as -2? or use -10? That's the only part I don't clearly understand because I could use any number, would it always give the correct answer?

So would I use:

-2 = p / ((20-p)/2)^2 · (p-20)/2

Sorry I didn't use tex, but it honestly wasn't working at all. It'd just show random signs...

 

[polak333]

So I tried it:

η = p / q · dq / dp
1 = p / q · dq / dp
1 = p/((20-p)/2)^2 · (p-20)/2
... (some work) ...
1 = (2p-40)/(20-p)^2
20 - p^2 = 2p - 40
0 = -60 + 2p + p^2
p = -8.81 (inadmissible)
p = 6.81

So therefore, 6.81 < p < 20.

For those values of p, η is elastic. Does it look right?

 

[Mentallic]

That's good. Now you could simplify the expression by noticing that (20-p)^2=(p-20)^2 and cancelling, but it's not completely necessary. In fact, for the way I would go about solving the next task in this problem, it's actually better in the form it already is.

If we are told that for some number x, |x|&gt;1 then this means that x&lt;-1, x&gt;1 because if we use say, x=-2 then |-2|=2>1.

Now you need to solve both inequalities for p. \eta&gt;1 and \eta&lt;-1 and find the intersection between the range of p that you found. For example, if you find for the case where \eta&gt;1 that 1&lt;p&lt;3 and in the second case you find that 2&lt;p&lt;4, both cases need to hold for |\eta|&gt;1 so the answer would be 2&lt;p&lt;3.

 

[polak333]

Ok, never mind the post above, it was totally wrong already on the second step.

Anyways, this time I think I got it:

-1 = p/((20-p)/2)^2 · (p-20)/2
-1 = (2p^2-40p)/(20-p)^2
-(20-p)^2 = 2p^2 - 40p
-p^2 + 40p - 400 = 2p^2 - 40p
0 = 3p^2 - 80p + 400

p = 20/3, 20

Did the same using:
1 = p/((20-p)/2)^2 · (p-20)/2
... (work) ...
p = +- 20

Therefore, 20/3 < p < 20.
I believe that's it, however they don't give an answer.

Now I can check by plugging in 20/3 and (something less than) 20 as η, and hopefully |η| > 1.

 

[Mentallic]

You should be solving inequalities -

\frac{2p(p-20)}{(p-20)^2}&lt;-1

and

\frac{2p(p-20)}{(p-20)^2}&gt;1

But anyway you still get the same answer (remember that solving inequalities requires that you multiply through by a positive number, or by a negative number and reversing the sign - which is why having a perfect square in the denominator was better than simplifying).

So your range for |\eta|&gt;1 is 20/3&lt;p&lt;20 so yes, plugging in any value of p in that range will give your desired result. Plugging in 20/3 or 20 will give |\eta|=1.

 

[polak333]

Thanks.

So for:
1 = p/((20-p)/2)^2 · (p-20)/2

it would be:
1 < p/((20-p)/2)^2 · (p-20)/2
... (work) ...
1 < (2p^2-40p)/((20-p)^2
p^2 - 40p + 400 < 2p^2 - 40p
0 < p^2 - 400

p > 20 or P < -20

Is that how it would look like?

 

[Mentallic]

Yep that's it

But notice that the demand equation is only valid for 0\leq p\leq 20 so you can scratch those values - which means you will never actually get \eta&gt;1.

 

[polak333]

Thanks!