Question

a university will do a seasonal run of a popular alumni sweatshirt. the revenue function is $r(x) = -0.006x^2 + 20x$, and the cost function is $c(x) = 1,600 + 7.40x$. in both functions, $x$ is the number of units produced. what is the larger number of sweatshirts the university can sell to break even? round to the nearest whole number. 1,050 units 1,964 units 5,015 units 0 units

Explanation:

Step1: Set revenue equal to cost

$R(x) = C(x)$
$-0.006x^2 + 20x = 1600 + 7.40x$

Step2: Rearrange to standard quadratic form

$-0.006x^2 + 20x - 7.40x - 1600 = 0$
$-0.006x^2 + 12.6x - 1600 = 0$
Multiply through by -1:
$0.006x^2 - 12.6x + 1600 = 0$

Step3: Apply quadratic formula

For $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Here $a=0.006$, $b=-12.6$, $c=1600$
First calculate discriminant:
$\Delta = (-12.6)^2 - 4(0.006)(1600) = 158.76 - 38.4 = 120.36$
$\sqrt{\Delta} = \sqrt{120.36} = 10.97$
Then:
$x=\frac{12.6\pm10.97}{2(0.006)}$

Step4: Calculate two solutions

First solution (using +):
$x=\frac{12.6+10.97}{0.012}=\frac{23.57}{0.012}\approx1964$
Second solution (using -):
$x=\frac{12.6-10.97}{0.012}=\frac{1.63}{0.012}\approx136$

Answer:

1,964 units