Suppose r(x) = 8(x)^(1/2) represents revenue and c(x) = 2(x)^2 represents cost, with x measured in thousands of units. Is there a production level that maximizes profit? If so, what is it?

profit = revenue - cost

= 8√x - 2x^2

d(profit)/dx = 4/√x - 4x
= 0 for a max of profit

4/√x = 4x
x = 1

since x was measured in thousands,
they should have a production of 1000

Ah, the thrilling world of numbers and profits! Let me clownishly calculate it for you.

To find the production level that maximizes profit, we need to determine the difference between revenue and cost, which is also known as profit. So, let's subtract the cost function from the revenue function:

p(x) = r(x) - c(x)
= 8√x - 2x^2

To find the x-value that maximizes profit, we need to find the critical points of the profit function p(x). This can be done by taking the derivative and setting it equal to zero:

p'(x) = (∂/∂x)(8√x - 2x^2)
p'(x) = 4/x^(1/2) - 4x

Now, we set p'(x) = 0 and solve for x:

0 = 4/x^(1/2) - 4x

By simplifying, we get:

4x = 4√x

Squaring both sides:

16x^2 = 16x

Dividing both sides by 16x:

x = 1

So, the production level that maximizes profit is x = 1 thousand units. Keep in mind, this is a simplified answer based on the given functions. But don't clown around too much with this answer; remember, there could be additional factors to consider in a real-world scenario!

To find the production level that maximizes profit, we need to determine the profit function first. The profit function (P) is given by subtracting the cost function (c(x)) from the revenue function (r(x)):

P(x) = r(x) - c(x)

Using the given revenue and cost functions, we can substitute their expressions:

P(x) = 8(x)^(1/2) - 2(x)^2

To find the production level that maximizes profit, we can find the critical points of the profit function.

To do this, we take the derivative of the profit function with respect to x:

P'(x) = (8/2)(1/2)(x)^(-1/2) - 2(2)(x)^1 = 4(x)^(-1/2) - 4x

Setting P'(x) equal to zero and solving for x, we get:

4(x)^(-1/2) - 4x = 0

Dividing both sides by 4 gives:

(x)^(-1/2) - x = 0

Rearranging the equation, we have:

x - (x)^(3/2) = 0

Factoring out x, we get:

x(1 - x^(1/2)) = 0

This equation has two solutions:

1) x = 0
2) 1 - x^0.5 = 0

Solving the second equation, we have:

1 - x^0.5 = 0
x^0.5 = 1
Taking the square of both sides, we get:
x = 1

Since x represents the number of thousands of units, x = 1 corresponds to a production level of 1000 units.

Therefore, the production level that maximizes profit is 1000 units.

To find the production level that maximizes profit, we need to determine the profit function first. The profit function can be obtained by subtracting the cost function from the revenue function.

Profit (P(x)) = Revenue (R(x)) - Cost (C(x))

In this case, the revenue function is given as r(x) = 8√x, and the cost function is given as c(x) = 2x^2.

Let's substitute these values into the profit function:

P(x) = r(x) - c(x)
P(x) = 8√x - 2x^2

Now, to find the production level that maximizes profit, we need to find the critical points of the profit function. Critical points occur where the derivative of the function is equal to zero or is undefined.

Let's take the derivative of the profit function:

P'(x) = d/dx [8√x - 2x^2]

Using the power rule and the chain rule, we can differentiate each term separately:

P'(x) = 4/x^(1/2) - 4x

Now, set P'(x) equal to zero and solve for x to find the critical point:

4/x^(1/2) - 4x = 0

Multiply both sides by x^(1/2):

4 - 4x^(3/2) = 0

Rearranging the equation, we have:

4x^(3/2) = 4

Divide both sides by 4:

x^(3/2) = 1

Take both sides to the power of 2/3:

(x^(3/2))^(2/3) = 1^(2/3)

x^(1) = 1

Simplifying, we have:

x = 1

So, the critical point in this case occurs at x = 1.

Now, we need to determine whether this critical point x = 1 corresponds to a maximum or minimum point. We can do this by examining the second derivative of the profit function.

Taking the second derivative:

P''(x) = d^2/dx^2 [4/x^(1/2) - 4x]
P''(x) = (3/2)x^(-3/2) - 4

Substitute x = 1 into the second derivative:

P''(1) = (3/2)(1)^(-3/2) - 4
P''(1) = (3/2)(1) - 4
P''(1) = 3/2 - 4
P''(1) = -5/2

Since the second derivative P''(1) is negative, it means that x = 1 corresponds to a maximum value.

Therefore, the production level that maximizes profit is x = 1 thousand units.