The weekly demand for the Pulsar 25-in. color console television is given by the demand equation

p = -0.03 x + 571\ \ \ \ \(0<=x<=12,000\)
where p denotes the wholesale unit price in dollars and x denotes the quantity demanded. The weekly total cost function associated with manufacturing these sets is given by
C(x) = 0.000002 x^3 - 0.01 x^2 + 400x + 80,000
where C(x) denotes the total cost incurred in producing x sets. Find the level of production that will yield a maximum profit for the manufacturer. Hint: Use the quadratic formula. (Round your answer to the nearest whole number.)
units

nobody has answered this yet, smh

To find the level of production that will yield a maximum profit for the manufacturer, we need to determine the profit equation and then calculate the maximum point using the quadratic formula.

The profit equation is given by the difference between the revenue and the cost:
Profit = Revenue - Cost

The revenue is obtained by multiplying the quantity demanded (x) by the unit price (p):
Revenue = x * p

Substituting the demand equation into the revenue equation, we have:
Revenue = x * (-0.03x + 571)
Revenue = -0.03x^2 + 571x

Now, we'll substitute the total cost function into the profit equation:
Profit = (-0.03x^2 + 571x) - (0.000002x^3 - 0.01x^2 + 400x + 80,000)
Profit = -0.000002x^3 + 0.01x^2 + 171x - 80,000

To find the level of production that maximizes the profit, we need to find the maximum point on the profit function. This can be done by finding the x-coordinate at the vertex of the profit function.

The profit function is a cubic equation, so let's first find the critical points on the graph by taking the derivative of the profit function and setting it equal to zero. Then we'll use the quadratic formula to solve for the x-coordinate of the vertex.

To find the derivative of the profit function:
Profit' = -0.000006x^2 + 0.02x + 171

Setting the derivative equal to zero and solving for x:
-0.000006x^2 + 0.02x + 171 = 0

This is a quadratic equation, so we can use the quadratic formula to find the x-coordinate of the vertex:
x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -0.000006, b = 0.02, and c = 171. Plugging these values into the formula, we get:
x = (-0.02 ± √(0.02^2 - 4(-0.000006)(171))) / (2(-0.000006))

Now we can calculate the x-coordinates:

x = (-0.02 ± √(0.0004 + 0.004104)) / (-0.000012)
x = (-0.02 ± √0.004504) / (-0.000012)
x = (-0.02 ± 0.067) / (-0.000012)

Now we have two possible x-coordinates for the vertex:

x₁ = (-0.02 + 0.067) / (-0.000012) = 4,583 (approx.)
x₂ = (-0.02 - 0.067) / (-0.000012) = 1,750 (approx.)

Next, evaluate the profit function at these two points to determine which one yields the maximum profit:

Profit at x₁ = -0.000002(4,583)^3 + 0.01(4,583)^2 + 171(4,583) - 80,000 = $1,041,096 (approx.)
Profit at x₂ = -0.000002(1,750)^3 + 0.01(1,750)^2 + 171(1,750) - 80,000 = $1,040,919 (approx.)

Comparing the profits at x₁ and x₂, we see that the maximum profit is achieved when x = 4,583 units.