Question

soccer ball profit
$y = -6x^2 + 100x - 180$
suppose the store wants to earn a daily profit of
$\\$150$ from the sale of soccer balls. to earn this
profit, what price should the store charge for each
soccer ball? explain how to solve this problem.

Explanation:

Step1: Set profit equal to 150

We know the profit function is \( y = -6x^{2}+100x - 180 \), and we want \( y = 150 \). So we set up the equation:
\( 150=-6x^{2}+100x - 180 \)

Step2: Rearrange to standard quadratic form

First, move all terms to one side to get a quadratic equation in the form \( ax^{2}+bx + c = 0 \).
Add \( 6x^{2}-100x + 180 \) to both sides:
\( 6x^{2}-100x + 330 = 0 \)
We can simplify this equation by dividing all terms by 2:
\( 3x^{2}-50x + 165 = 0 \)

Step3: Solve the quadratic equation

We can use the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 3 \), \( b=-50 \), and \( c = 165 \).
First, calculate the discriminant \( D=b^{2}-4ac=(-50)^{2}-4\times3\times165=2500 - 1980 = 520 \)
Then, \( x=\frac{50\pm\sqrt{520}}{6}=\frac{50\pm2\sqrt{130}}{6}=\frac{25\pm\sqrt{130}}{3} \)
Calculate the approximate values:
\( \sqrt{130}\approx11.40 \)
\( x_1=\frac{25 + 11.40}{3}=\frac{36.40}{3}\approx12.13 \)
\( x_2=\frac{25 - 11.40}{3}=\frac{13.60}{3}\approx4.53 \)

Answer:

The store can charge approximately \(\$4.53\) or \(\$12.13\) for each soccer ball to earn a daily profit of \(\$150\). (If we want to present the exact form, the solutions are \( x=\frac{25+\sqrt{130}}{3} \) and \( x=\frac{25 - \sqrt{130}}{3} \))