The revenue from the sale of a product is, in dollars,

R = 1500x + 3000(2x + 3)^−1 − 1000
where x is the number of units sold. Find the marginal revenue when 150 units are sold. (Round your answer to two decimal places.)

Oh, revenue calculations! Time to put on my accountant wig and crunch some numbers.

To find the marginal revenue when 150 units are sold, we need to calculate the derivative of the revenue equation with respect to x.

So, let's get started!

First, let's find the derivative of each term.

The derivative of 1500x is simply 1500.

The derivative of 3000(2x + 3)^-1 is a little trickier. We can use the chain rule, which states that the derivative of a function within another function can be found by multiplying the derivative of the inner function by the derivative of the outer function.

Using the chain rule, the derivative of (2x + 3)^-1 is -2/(2x + 3)^2, and then we multiply that by the derivative of (2x + 3), which is 2. This gives us a derivative of -4/(2x + 3)^2.

Since -1000 is a constant, its derivative is simply zero.

Now, let's add up these derivatives to find the marginal revenue.

Marginal Revenue = 1500 + (-4/(2x + 3)^2)

Plugging in x = 150, we get

Marginal Revenue = 1500 + (-4/(2(150) + 3)^2)

Simplifying this, we have:

Marginal Revenue = 1500 + (-4/(2(150) + 3)^2)

Marginal Revenue = 1500 + (-4/303)^2

Marginal Revenue ≈ 1500 - 0.0004408

So, the marginal revenue when 150 units are sold is approximately $1500 - $0.0004408.

To find the marginal revenue when 150 units are sold, we need to compute the derivative of the revenue function with respect to x and evaluate it at x = 150.

Taking the derivative of the revenue function R = 1500x + 3000(2x + 3)^-1 - 1000, we have:

dR/dx = 1500 - 3000(2x + 3)^-2 * 2

Simplifying this expression, we get:

dR/dx = 1500 - (12000 / (2x + 3)^2)

Now, substitute x = 150 into this derivative equation:

dR/dx = 1500 - (12000 / (2(150) + 3)^2)

dR/dx = 1500 - (12000 / (300 + 3)^2)

dR/dx = 1500 - (12000 / 903^2)

Calculating the value, we find:

dR/dx ≈ 1500 - (12000 / 815409)

dR/dx ≈ 1500 - 0.0147

dR/dx ≈ 1499.99

Therefore, the marginal revenue when 150 units are sold is approximately $1499.99.

To find the marginal revenue when 150 units are sold, we need to find the derivative of the revenue function with respect to x, and then evaluate it at x = 150.

First, let's find the derivative of the revenue function R(x) = 1500x + 3000(2x + 3)^-1 - 1000.

To find the derivative of x, we need to apply the power rule and the chain rule.

The power rule states that the derivative of x^n is nx^(n-1).

The chain rule states that the derivative of f(g(x)) is f'(g(x)) * g'(x), where f(x) is the outer function and g(x) is the inner function.

Now let's find the derivative step by step:

R'(x) = 1500 * (d/dx) x + 3000 * (d/dx) (2x + 3)^-1 - 0

The derivative of x is 1.

For (2x + 3)^-1, we can use the chain rule.

Let u = 2x + 3.

Then, R'(x) = 1500 * 1 + 3000 * (d/du) u^-1 * (d/dx) (2x + 3)

= 1500 + 3000 * (-1/u^2) * 2

= 1500 - 6000/u^2

Now, let's substitute u back in.

u = 2x + 3

R'(x) = 1500 - 6000/(2x + 3)^2

Now we have the derivative of the revenue function, which represents the marginal revenue.

To find the marginal revenue when 150 units are sold (x = 150), we substitute x = 150 into R'(x) and evaluate it.

R'(150) = 1500 - 6000/(2 * 150 + 3)^2

R'(150) = 1500 - 6000/(300 + 3)^2

R'(150) ≈ 1500 - 6000/(303)^2

R'(150) ≈ 1500 - 6000/91809

R'(150) ≈ 1500 - 0.06529

R'(150) ≈ 1499.93

Therefore, the marginal revenue when 150 units are sold is approximately $1499.93.

dR/dx = 1500 - 3000*2/(2x+3)^2

if x = 150

dR/dx = 1500 - 6000/303^2

= $1499.93 per unit

the point is that you have an initial 1000 setup cost
then it is $1500 per unit but that decreases as the number goes up and you train robots