Question

a business owner is trying to determine the break-even point. the owner uses the revenue function $r(x) = -0.012x^2 + 800x$. the cost function includes an initial investment of $5,000, along with $550 per unit. write a linear cost function where $x$ is the number of units sold, and use the functions to find the larger number of units the company can sell to break even. round to the nearest thousand. enter your answer in the box. units

Explanation:

Step1: Define Cost Function

The cost function \( C(x) \) has a fixed cost of $5000 and variable cost of $550 per unit, so \( C(x) = 5000 + 550x \).

Step2: Set Revenue = Cost

At break - even, \( R(x)=C(x) \). So, \( - 0.012x^{2}+800x=5000 + 550x \).

Step3: Rearrange to Quadratic Equation

Rearrange the equation: \( - 0.012x^{2}+800x-550x - 5000 = 0 \), which simplifies to \( -0.012x^{2}+250x - 5000=0 \). Multiply through by - 1 to get \( 0.012x^{2}-250x + 5000 = 0 \).

Step4: Use Quadratic Formula

For a quadratic equation \( ax^{2}+bx + c = 0 \), the solutions are \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). Here, \( a = 0.012 \), \( b=-250 \), \( c = 5000 \).

First, calculate the discriminant \( \Delta=b^{2}-4ac=(-250)^{2}-4\times0.012\times5000=62500 - 240 = 62260 \).

Then, \( x=\frac{250\pm\sqrt{62260}}{2\times0.012}=\frac{250\pm249.52}{0.024} \).

We have two solutions:
\( x_1=\frac{250 + 249.52}{0.024}=\frac{499.52}{0.024}\approx20813.33 \)
\( x_2=\frac{250 - 249.52}{0.024}=\frac{0.48}{0.024} = 20 \)

Step5: Round the Larger Solution

The larger number of units is approximately 21000 (when rounded to the nearest thousand).

Answer:

21000