A software company's weekly revenue, R, in thousands of dollars, is modeled by the functio… — ACT

A software company's weekly revenue, $R$ , in thousands of dollars, is modeled by the function $R(x) = -x^2 + 8x$ , where $x$ represents the price of the software in dollars. The company's weekly operating cost, $C$ , in thousands of dollars, is modeled by the function $C(x) = 2x + 5$ . The company breaks even when its weekly revenue equals its weekly operating cost. What is the sum of the possible software prices, in dollars, at which the company breaks even?

Answer: 6 dollars

Answer

The sum of the possible software prices at which the company breaks even is 6.

To find the software prices at which the company breaks even, we set the revenue function

R(x)

equal to the cost function

C(x)

, which gives

-x^2 + 8x = 2x + 5

. Rearranging this equation into standard quadratic form gives

x^2 - 6x + 5 = 0

. Factoring this expression yields

(x - 5)(x - 1) = 0

, meaning the individual solutions are

x = 5

and

x = 1

. The sum of these possible break-even software prices is

5 + 1 = 6

Step-by-Step Solution

Set the revenue function equal to the cost function.

-x^2 + 8x = 2x + 5

The break-even points occur where the revenue and cost are equal.

Rearrange the terms to set the equation to zero.

x^2 - 6x + 5 = 0

Moving all terms to one side of the equation puts it in standard quadratic form, allowing us to solve for

x

Factor the quadratic equation.

(x - 5)(x - 1) = 0

Factoring is a standard method to find the roots of a quadratic equation of the form

ax^2 + bx + c = 0

Find the individual roots by setting each factor to zero.

x = 5

and

x = 1

If the product of two factors is zero, at least one of the factors must be zero.

Calculate the sum of the solutions.

5 + 1 = 6

The question specifically asks for the sum of the possible prices.

Key Concept

Solving systems of linear and non-linear equations by setting the equations equal to each other and solving the resulting quadratic equation.

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