MATH SOLVE

4 months ago

Q:
# 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?

Accepted Solution

A:

Answer:a) [tex]\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)[/tex]b) -0.50c) $7.50Step-by-step explanation:a)
The price elasticity of demand E at a given point [tex]\large (p_1,q_1)[/tex] is defined as
[tex]\large E=\displaystyle\frac{p_1}{q_1}.\displaystyle\frac{\text{d}q}{\text{d}p}(p_1)[/tex]
and it is supposed to measure the possible response of tissues demand due to small changes in its price when the price is at [tex]\large p_1[/tex]
(a) Obtain a formula for the price elasticity of demand for E = mc2 T-shirts.
Since
[tex]\large \displaystyle\frac{\text{d}q}{\text{d}p}=-4p+30,\;(9<p<15)[/tex]
[tex]\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)[/tex]
where [tex]\large 9<p_1<15[/tex]
(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
[tex]\large q=-2(9)^2+30(9)=108[/tex]
t-shirts
[tex]\large (p_1,q_1)=(9,108)\Rightarrow \frac{p_1}{q_1}=\displaystyle\frac{9}{108}=0.0833[/tex]
Also, we have
[tex]\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[/tex]
hence
[tex]\large E=\displaystyle\frac{p_1}{q_1}.\displaystyle\frac{\text{d}q}{\text{d}p}(p_1)=0.0833(-6)=-0.50[/tex]
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 [tex]\large \displaystyle\frac{\text{d}q}{\text{d}p}=-4p+30[/tex]
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
[tex]\large q=-2(7.5)^2+30=86.25 \approx 86[/tex] T-shirts.