Question

Suppose the demand function for x thousand of a certain item is p=100+70/ln x, where x>1 and p is in dollars. (a) Find the marginal revenue. (b) Find the revenue from the next thousand items at a demand of 4000(x=4).

Answer 1

The marginal revenue is the derivative of the revenue function, which is obtained by multiplying the demand function by quantity. Once the marginal revenue function is derived, the revenue from the next thousand items at a demand of 4000 can be found by evaluating that function at x=4.

Step-by-step explanation:

To address part (a) of the question, the marginal revenue can be found by first calculating the revenue function through the demand function, where revenue (R) is the product of price (p) and quantity (x), leading to R(x) = x × (100 + 70/ln(x)). The marginal revenue is then the derivative of the revenue function with respect to x, MR(x) = dR/dx. To find this derivative, we must apply product rule and the chain rule of differentiation.

For part (b), once the marginal revenue function is found, to find the revenue from the next thousand items when the demand is 4000 (x=4), we plug x=4 into the marginal revenue function and find MR(4).

Answer 2

(a) Marginal revenue:
`\(MR = 100 - (70)/((\ln x)^2) \cdot (1)/(x)\)`

(b) Revenue from the next thousand items at
`\(x = 4\): \(R_{\text{next thousand}} = 4 \cdot \left(100 + (70)/(\ln 4)\right) - 3 \cdot \left(100 + (70)/(\ln 3)\right)\)`

To find the marginal revenue and the revenue from the next thousand items given the demand function
`\(p = 100 + (70)/(\ln x)\)`, we'll proceed step by step.

(a) Find the Marginal Revenue:

The derivative of the revenue function with respect to the quantity sold is known as the marginal revenue (MR).

The revenue function (R) is found by multiplying the demand function by the quantity sold, (x):

`\[ R = x \cdot p \]`

Given
`\(p = 100 + (70)/(\ln x)\)`, we can express the revenue function in terms of x:

`\[ R = x \cdot p = x \left(100 + (70)/(\ln x)\right) \]`

Take the revenue function's derivative with respect to x to determine the marginal revenue:

`\[ (dR)/(dx) = (d)/(dx) \left(x \cdot p\right) = 100 - (70)/((\ln x)^2) \cdot (1)/(x) \]`

This derivative represents the marginal revenue function.

(b) Find the Revenue from the Next Thousand Items at a Demand of 4000 (x = 4):

At x = 4, we want to find the revenue from the next thousand items. To find this, first, calculate the revenue for 4000 items (x = 4):

`\[ R = x \cdot p = 4 \cdot \left(100 + (70)/(\ln 4)\right) \]`

Then, find the revenue for 3000 items (x = 3) to determine the revenue from the next thousand items:

`\[ R_{\text{next thousand}} = R(4000) - R(3000) \]`

Solving these expressions will give us the revenue from the next thousand items at a demand of 4000.

If you'd like, I can help you calculate these derivatives and the specific revenue values for x = 4 and x = 3 to determine the revenue from the next thousand items.