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) Â 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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