Respuesta :
Answer with Step-by-step explanation:
Since we have given that
q = 896-20p
p = $32
.(A) Calculate the price elasticity of demand
As we know that
[tex]e=-\dfrac{dq}{dp}\times \dfrac{p}{q}\\\\e=-(-20)\times \dfrac{32}{896-20p}\\\\e=\dfrac{20\times 32}{896-20p}\\\\e=\dfrac{640}{896-20p}\\\\e=\dfrac{640}{896-640}\\\\e=\dfrac{640}{256}=2.5[/tex]
(B) The demand is going down with increase in 15 increase in price at that price level, as we know that there is inverse relationship between price and quantity demanded.
(C) Also, calculate the price that gives a maximum weekly revenue.
[tex]R=pq\\\\R=p(896-20p)\\\\R=896p-20p^2[/tex]
We first find the first derivative:
[tex]R'(p)=896-400p<0[/tex]
So, it becomes,
[tex]896-40p=0\\\\40p=896\\\\p=\dfrac{896}{40}=\$22.4[/tex]
R=-40<0, so, it will give maximum revenue.
(D) Find this maximum revenue.
Maximum revenue would be [tex]R=pq\\\\R=22.4(896-20\times 22.4)\\\\R=\$10035.2[/tex]
The price elasticity of demand is simply the change in the quantity purchased relative to a change in price.
- The price elasticity of demand is 2.5
- The price that gives maximum revenue is $22.4
- The maximum revenue is $10035.2
The given parameters are:
[tex]\mathbf{q = 896 - 20p}[/tex]
(a) Price elasticity of demand, when p = 32
This is calculated using:
[tex]\mathbf{e = -\frac{dp}{dq} \times \frac pq}[/tex]
For a quantity function:
[tex]\mathbf{q = mp + c}[/tex]
[tex]\mathbf{\frac{dp}{dq} = m}[/tex]
So, by comparison:
[tex]\mathbf{\frac{dp}{dq} = -20}[/tex]
So, we have:
[tex]\mathbf{e = -\frac{dp}{dq} \times \frac pq}[/tex]
[tex]\mathbf{e = -(-20) \times \frac{32}{896 -20(32)}}[/tex]
[tex]\mathbf{e = -(-20) \times \frac{32}{256}}[/tex]
[tex]\mathbf{e = \frac{640}{256}}[/tex]
[tex]\mathbf{e = 2.5}[/tex]
The price elasticity of demand is 2.5
(b) Interpret the price elasticity
The demand is going ___down____ by ___1.5%___% per 1% increase in price at that price level
(c) The maximum weekly revenue
This is calculated as:
[tex]\mathbf{r = pq}[/tex]
So, we have:
[tex]\mathbf{r = p(896 - 20p)}[/tex]
Open bracket
[tex]\mathbf{r = 896p - 20p^2}[/tex]
Differentiate
[tex]\mathbf{r = 896 - 40p}[/tex]
Set to 0
[tex]\mathbf{896 - 40p = 0}[/tex]
Collect like terms
[tex]\mathbf{40p = 896 }[/tex]
Divide both sides by 40
[tex]\mathbf{p = 22.4 }[/tex]
Hence, the price that gives maximum revenue is $22.4
(d) The maximum revenue
In (c), we have:
[tex]\mathbf{r = p(896 - 20p)}[/tex]
Substitute 22.4 for p
[tex]\mathbf{r = 22.4 \times (896 - 20 \times 22.4)}[/tex]
[tex]\mathbf{r = 10035.2}[/tex]
Hence, the maximum revenue is $10035.2
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