Problem

Given the function below, complete parts (a) through (c).
$q=D(x)=\frac{k}{x^{n}}$, where $k$ is a positive constant and $n$ is an integer greater than 0 .
(a) Find the elasticity of the demand function.
(b) Is the value of the elasticity dependent on the price per unit?
(c) Does the total revenue have a maximum? When?
(a) $E(x)=$
(Simplify your answer.)

Answer

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Answer

Final Answer: (c) The total revenue does not have a maximum.

Steps

Step 1 :The elasticity of a function is given by the formula \(E(x) = x \cdot \frac{f'(x)}{f(x)}\). In this case, \(f(x) = \frac{k}{x^n}\), so we need to find the derivative of \(f(x)\) first.

Step 2 :The derivative of \(f(x)\) with respect to \(x\) can be found using the power rule for differentiation, which states that the derivative of \(x^n\) is \(n \cdot x^{n-1}\). Therefore, the derivative of \(f(x)\) is \(f'(x) = -n \cdot \frac{k}{x^{n+1}}\).

Step 3 :We can then substitute \(f(x)\) and \(f'(x)\) into the formula for \(E(x)\) to find the elasticity of the function. The elasticity of the function is \(-n\), which is a constant. This means that the elasticity does not depend on the value of \(x\), which in this context represents the price per unit.

Step 4 :Final Answer: (a) The elasticity of the demand function is \(\boxed{-n}\). (b) The value of the elasticity is not dependent on the price per unit.

Step 5 :For part (c), we need to find the total revenue function and determine whether it has a maximum. The total revenue \(R(x)\) is given by the product of the price per unit \(x\) and the demand \(D(x)\), i.e., \(R(x) = x \cdot D(x) = x \cdot \frac{k}{x^n} = \frac{k}{x^{n-1}}\).

Step 6 :To find whether \(R(x)\) has a maximum, we can take its derivative and set it equal to zero. If there exists a value of \(x\) that satisfies this equation, then \(R(x)\) has a maximum at that value of \(x\).

Step 7 :The derivative of the total revenue function is \(R'(x) = -k \cdot n \cdot x^{-n} + k \cdot x^{-n}\). Setting this equal to zero gives the equation \(-k \cdot n \cdot x^{-n} + k \cdot x^{-n} = 0\). However, this equation has no solutions for \(x\), which means that the total revenue function does not have a maximum.

Step 8 :Final Answer: (c) The total revenue does not have a maximum.

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