1. Elasticity

The price elasticity is given by

$\displaystyle \epsilon = -\frac{p}{q}\cdot\frac{dq}{dp}$

and the demand equation is given by
$\displaystyle q = 700-35p$
and p is given as 4 for $\displaystyle 0\leq p \leq 20$

So i plus in 4. Which gives q as 560.

Then i take derivative of q which is 35? and derivative of p which is one?

Then throw them in the elasticity equation and get -0.25.

Does that make sense?

2. The derivative of $\displaystyle q$ is -35, and you don't need the derivative of $\displaystyle p$; $\displaystyle p = 4.$

3. so i have

$\displaystyle \frac{4}{560}\cdot\frac{-35}{0}$

That cant be right.

if $\displaystyle p= \frac{700-q}{35}$

then derivative of p is one?

4. Hi el123

Originally Posted by el123
so i have

$\displaystyle \frac{4}{560}\cdot\frac{-35}{0}$

That cant be right.

if $\displaystyle p= \frac{700-q}{35}$

then derivative of p is one?
According to BobP's statement

it is
$\displaystyle \epsilon = -\frac{p}{q}\cdot\frac{dq}{dp}$

and the demand equation is given by
$\displaystyle q = 700-35p$
$\displaystyle \epsilon = - \frac{p}{q}*\frac{dq}{dp}$

and

$\displaystyle q = 700-35p$

=> $\displaystyle \frac{dq}{dp} = y' = -35$

so $\displaystyle \epsilon = - \frac{p}{q}*\frac{dq}{dp} = - \frac{p}{q}*(-35)$

Yours
Rapha

5. If $\displaystyle q = 700 -35p,$, then the derivative of $\displaystyle q$ wrt $\displaystyle p$ is

$\displaystyle \frac{dq}{dp} = -35.$

Also, if $\displaystyle q = 700 - 35p,$ then $\displaystyle p = \frac{700 - q}{35}$

and the derivative of $\displaystyle p$ wrt $\displaystyle q$ would be

$\displaystyle \frac{dp}{dq} = -\frac{1}{35}$,

but that particular derivative is not needed.

6. indeed you are right.I see my mistake.

Thanks guys ...or girls.

One more thing , what does the ...for $\displaystyle 0 \leq p\leq 20$ mean?

7. It means that p is to be greater than or equal to zero, and less than or equal to 20.

i.e. p has to lie within the range zero to 20 inclusive.