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, $$|x|>1$$ then this means that $$x<-1, x>1$$ because if we use say, x=-2 then |-2|=2>1.

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

 

[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 -

$$\frac{2p(p-20)}{(p-20)^2}<-1$$

and

$$\frac{2p(p-20)}{(p-20)^2}>1$$

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 $$|\eta|>1$$ is $$20/3<p<20$$ 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 $$\eta>1$$.

 

[polak333]

Thanks!