Question

The Physics Club sells E = mc2 T-shirts at the local flea market. Unfortunately, the club's previous administration has been losing money for years, so you decide to do an analysis of the sales. A quadratic regression based on old sales data reveals the following demand equation for the T-shirts: q = −2p2 + 30p (9 ≤ p ≤ 15). Here, p is the price the club charges per T-shirt, and q is the number it can sell each day at the flea market.
(a) Obtain a formula for the price elasticity of demand for E = mc2 T-shirts.
(b) Compute the elasticity of demand if the price is set at $9 per shirt. (Round your answer to two decimal places.) Interpret the result.
(c) How much should the Physics Club charge for the T-shirts to obtain the maximum daily revenue? What will the revenue be?

Answer 1

Answer:

a) $\large E=\displaystyle\frac{p_1}{q_1}.\displaystyle\frac{\text{d}q}{\text{d}p}(p_1)=\displaystyle\frac{p_1}{q_1}(-4p_1+30)$

b) -0.50

c) $7.50

Step-by-step explanation:

a)

The price elasticity of demand E at a given point $\large (p_1,q_1)$ is defined as

$\large E=\displaystyle\frac{p_1}{q_1}.\displaystyle\frac{\text{d}q}{\text{d}p}(p_1)$

and it is supposed to measure the possible response of tissues demand due to small changes in its price when the price is at $\large p_1$

(a) Obtain a formula for the price elasticity of demand for E = mc2 T-shirts.

Since

$\large \displaystyle\frac{\text{d}q}{\text{d}p}=-4p+30,\;(9<p<15)$

$\large E=\displaystyle\frac{p_1}{q_1}.\displaystyle\frac{\text{d}q}{\text{d}p}(p_1)=\displaystyle\frac{p_1}{q_1}(-4p_1+30)$

where $\large 9<p_1<15$

(b) Compute the elasticity of demand if the price is set at $9 per shirt. (Round your answer to two decimal places.) Interpret the result.

When the price is set at $9 the demand is

$\large q=-2(9)^2+30(9)=108$

t-shirts

$\large (p_1,q_1)=(9,108)\Rightarrow \frac{p_1}{q_1}=\displaystyle\frac{9}{108}=0.0833$

Also, we have

$\large \displaystyle\frac{\text{d}q}{\text{d}p}=-4p+30\Rightarrow \displaystyle\frac{\text{d}q}{\text{d}p}(9)=-4*9+30=-6$

hence

$\large E=\displaystyle\frac{p_1}{q_1}.\displaystyle\frac{\text{d}q}{\text{d}p}(p_1)=0.0833(-6)=-0.50$

That would mean the demand would go down about 0.5% per 1% increase in price at that price level.

(c) How much should the Physics Club charge for the T-shirts to obtain the maximum daily revenue? What will the revenue be?

Let T be the price at which the T-shirts are sold.

The daily revenue is Tq so Tq is maximum when q is maximum.

since  

$\large \displaystyle\frac{\text{d}q}{\text{d}p}=-4p+30$

there is a critical point when -4p + 30 = 0

and p should be (30/4) = $7.5

given that the second derivative of q with respect to p equals -4 < 0, that point is a maximum.

At that price, the demand would be

$\large q=-2(7.5)^2+30=86.25 \approx 86$ T-shirts.