Question

The profit for a certain commodity, n, where n is in units, is given by the functionP(n) =25n^2 + 300n + 1125At the break-even point, the profit is zero, i.e., P(n) = 0. Find the number of units where the break-even point is located, i.e., find n when P(n) = 0.The number of units, n, where the break-even point is located, isHint: Did you factor the GCF? Then reduce your equation to smaller coefficients?

Answer 1

Given the function:

$P\mleft(n\mright)=-25n^2+300n+1125$

You know that "P" is the profit for a certain "n" commodity (in units).

You need to find the value of "n" when:

$P(n)=0$

In order to find it, you can follow these steps:

1. Substitute this value into the function:

$P(n)=0$

Then:

$\begin{gathered} 0=-25n^2+300n+1125 \\ \\ -25n^2+300n+1125=0 \end{gathered}$

2. Divide both sides of the equation by -25:

$\begin{gathered} \frac{-25n^2}{(-25)}+\frac{300n}{(-25)}+\frac{1125}{(-25)}=\frac{0}{(-25)} \\ \\ n^2-12n-45=0 \end{gathered}$

3. Factor the equation by finding two numbers whose Sum is -12 and whose Product is -45. These are 3 and -15, because:

$\begin{gathered} 3-15=-12 \\ \\ (3)(-15)=-45 \end{gathered}$

Then:

$(n+3)(n-15)=0$

4. Set up these two equations:

$\begin{gathered} n+3=0\text{ (Equation 1)} \\ \\ n-15=0\text{ (Equation 2)} \end{gathered}$

5. Solving form "n" from each equation, you get:

$\begin{gathered} n+3=0\text{ }\Rightarrow n_1=-3 \\ \\ n-15=0\Rightarrow n_2=15 \end{gathered}$

Since the number of units cannot be negative, the answer is:

$n=15$