Questions: The demand for a product is given by x=F(p)=1840/(5+ln (2+p)). If p=30, determine E and interpret the results.

Transcript text: The demand for a product is given by $x=F(p)=\frac{1840}{5+\ln (2+p)}$. If $p=30$, determine $E$ and interpret the results.

Solution

Solution Steps

To solve this problem, we need to calculate the elasticity of demand $ E $ at a given price $ p = 30 $. The elasticity of demand is given by the formula $ E = \frac{p}{x} \cdot \frac{dx}{dp} $. First, we need to find $ x $ by substituting $ p = 30 $ into the demand function. Then, we calculate the derivative $ \frac{dx}{dp} $ of the demand function. Finally, we substitute these values into the elasticity formula to find $ E $.

Step 1: Calculate Demand at $ p = 30 $

We start by substituting $ p = 30 $ into the demand function $ x = \frac{1840}{5 + \ln(2 + p)} $:

\[ x = \frac{1840}{5 + \ln(32)} \approx \frac{1840}{5 + 3.465736} \approx \frac{1840}{8.465736} \approx 217.4 \]

Step 2: Calculate the Derivative $ \frac{dx}{dp} $

Next, we find the derivative of the demand function with respect to $ p $:

\[ \frac{dx}{dp} = -\frac{1840}{(p + 2)(5 + \ln(2 + p))^2} \]

Substituting $ p = 30 $:

\[ \frac{dx}{dp} = -\frac{1840}{(32)(5 + \ln(32))^2} \approx -\frac{1840}{32 \cdot (8.465736)^2} \approx -\frac{1840}{32 \cdot 71.703} \approx -\frac{1840}{2294.464} \approx -0.802 \]

Step 3: Calculate Elasticity $ E $

Now we can calculate the elasticity of demand $ E $ using the formula:

\[ E = \frac{p}{x} \cdot \frac{dx}{dp} \]

Substituting the values we found:

\[ E = \frac{30}{217.4} \cdot (-0.802) \approx -0.1107 \]