For the following demand function, find elasticity and the values of q at which total revenue is maximized. q=37,700-7p^2
To find the elasticity of demand, we can use the formula:
\[E = \frac{dq}{dp} \times \frac{p}{q}\]
First, find the derivative of the demand function with respect to price:
\[\frac{dq}{dp} = -14p\]
Next, plug in the values of q and p into the demand function to find the corresponding values:
\[q = 37,700 - 7p^2\]
When q = 0, p = √(5371.43)
Therefore, when q = 37,700, p = 0
Now, we can find the elasticity of demand at q = 37,700:
\[E = -14 \times 0 \times \frac{0}{37,700} = 0\]
The elasticity of demand at q = 37,700 is 0.
To find the values of q at which total revenue is maximized, we can use the formula for total revenue:
\[TR = q \times p\]
Substitute the demand function into the total revenue formula to get the new formula:
\[TR = (37,700 - 7p^2) \times p\]
\[TR = 37,700p - 7p^3\]
Next, find the derivative of the total revenue function with respect to p, set it equal to zero, and solve for q:
\[\frac{dTR}{dp} = 37,700 - 21p^2 = 0\]
\[21p^2 = 37,700\]
\[p^2 = 1790.48\]
\[p = 42.29\]
Plug the value of p back into the demand function to find q:
\[q = 37,700 - 7(42.29)^2\]
\[q = 18,850\]
The values of q at which total revenue is maximized are q = 18,850 and p = 42.29.