Question

Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is p dollars, the revenue R (in dollars) is R(p) = -5p^2 + 10,000p. What unit price should be established for the dryer to maximize revenue? What is the maximum revenue?

The unit price that should be established to maximize revenue is $___?
The maximum revenue is $___?

Answer 1

The unit price that should be established to maximize revenue is $1,000. The maximum revenue is $5,000,000.

Step-by-step explanation:

To find the unit price that maximizes revenue, we need to determine the vertex of the quadratic function R(p) = -5p^2 + 10,000p.

The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the x-coordinate:

x = -rac{b}{2a}

In this case, a = -5 and b = 10,000. Plugging these values into the formula, we get:

x = -rac{10,000}{2(-5)} = 1,000

Therefore, the unit price that should be established to maximize revenue is $1,000.

To find the maximum revenue, substitute the x-coordinate of the vertex back into the function:

R(1,000) = -5(1,000)^2 + 10,000(1,000)

R(1,000) = -5,000,000 + 10,000,000 = $5,000,000

Therefore, the maximum revenue is $5,000,000.

Answer 2

To maximize revenue for the gas clothes dryer, the unit price should be set at $1000, resulting in a maximum revenue of $5,000,000.

Step-by-step explanation:

The manufacturer of a gas clothes dryer discovers their revenue function as R(p) = -5p2 + 10,000p. To maximize revenue, we must first find the vertex of this quadratic function, where the axis of symmetry provides the value of p for maximum revenue. The axis of symmetry in a standard parabola given by y = ax2 + bx + c is x = -b/2a. For this revenue function, a = -5 and b = 10,000, so p = -10,000 / (2 * -5) = 1000. Hence, the unit price to maximize revenue is $1000.

Substitute p = 1000 into the revenue function to find the maximum revenue: R(1000) = -5(1000)2 + 10,000(1000) = -5,000,000 + 10,000,000 = $5,000,000. Therefore, the maximum revenue is $5,000,000.