Question

CALCULUS:

Recall that the elasticity of demand is defined by;

E(p) = $\frac{-p f'(p)}{f(p)}$


Let the demand equation be given by;

f(p)= $\sqrt[]{40-p^{2} }$


1. FIND E(p) and the compute E(3)
2. If E(p)=1 , what would this mean for the demand of the function?

Answer 1

Answer:

1.

$E(p)=\frac{p^2}{40-p^2}$

E(3) = 0.29

2. See Below

Step-by-step explanation:

1.

According to the formula we would need the derivative, so lets first calculate the derivative of f(p) given.

The rules to use would be $f(x)=ax^n\\f'(x)=nax^{n-1}$

Also, remember to use chain rule if there is a function inside a function. So we differentiate the "inside" function again.

Now, lets differentiate:

$f(p)=\sqrt{40-p^2}\\f(p)=(40-p^2)^{\frac{1}{2}}\\f'(p)=\frac{1}{2}(40-p^2)^{\frac{1}{2}-1}*(-2p)\\f'(p)=\frac{1}{2}(40-p^2)^{-\frac{1}{2}}*(-2p)\\f'(p)=-\frac{p}{\sqrt{40-p^2}}$

Now, the E(p):

$E(p)=-\frac{pf'(p)}{f(p)}\\E(p)=-\frac{p*(-\frac{p}{\sqrt{40-p^2}})}{\sqrt{40-p^2}}\\E(p)=\frac{p^2}{40-p^2}$

We find E(3) by substituting 3 into "p":

$E(p)=\frac{p^2}{40-p^2}\\E(3)=\frac{3^2}{40-3^2}\\E(3)=\frac{9}{31}\\E(3)=0.29$

2.

When E(p) = 1, it means the price elasticity of demand is "1". It is the called Unitary Elastic Demand. It means that for every unit change in price, there is the same unit change in demand. The changes of demand and price are proportional to each other.