Answer:
11
Step-by-step explanation:
Ok so we aren't looking to maximize profits which is equivalent to finding the vertex of the parabola.
But we want to know what is the highest number of x so that our we aren't losing money. We aren't losing money when our profit is 0.
So we are asked to solve P(x)=0 for x and take the highest x of that equation.
[tex]P(x)=0[/tex]
[tex]-10x^2+120x-150=0[/tex]
I'm going to divide both sides by -10 to make the numbers easier to work with later.
[tex]x^2-12x+15=0[/tex]
I'm going to use the quadratic formula.
[tex]x^2-12x+15=0[/tex] when compared to [tex]ax^2+bx+c=0[/tex] gives us [tex]a=1,b=-12,c=15[/tex].
We are going to plug this into our quadratic formula:
[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex].
[tex]x=\frac{12\pm \sqrt{(-12)^2-4(1)(15)}}{2(1)}[/tex]
[tex]x=\frac{12\pm \sqrt{144-60}}{2}[/tex]
[tex]x=\frac{12\pm \sqrt{84}}{2}[/tex]
[tex]x=\frac{12}{2} \pm \frac{\sqrt{84}}{2}[/tex]
[tex]x=6 \pm \frac{\sqrt{84}}{2}[/tex]
I'm ready to put this into my calculator.
The one with the plus gives me 10.5826 approximately.
The one with the minus gives me 1.4174 approximately.
The 10.5826 is the higher number.
So about 11 dollars.
Answer:
C $11
Step-by-step explanation:
If a company found that its monthly profit, P, is given by P = -10x2 + 120x - 150 where x is the selling price for each unit of the product, the best estimate of the maximum price per unit that the company can charge without losing money is $11.
