Question

Noah has a summer tree-trimming business. Based on experience, Noah knows that his profit, P, in dollars, can be modelled by = −3^2 + 150 − 1200, where x is the amount he charges per tree.
1. How much does he need to charge if he wants to break even?

2. How much does he need to charge if he wants to make a profit of $600?

Answer 1

Solving quadratic equations, it is found that he needs to charge:

1. He needs to charge $40 to break even.

2. He needs to charge $30 for a profit of $600.

What is a quadratic function?

A quadratic function is given according to the following rule:

$y = ax^2 + bx + c$

The solutions are:

• $x_1 = \frac{-b + \sqrt{\Delta}}{2a}$

• $x_2 = \frac{-b - \sqrt{\Delta}}{2a}$

In which:

$\Delta = b^2 - 4ac$

The profit equation in this problem is:

P(x) = -3x² + 150x - 1200.

He breaks even when P(x) = 0, hence:

-3x² + 150x - 1200 = 0.

The coefficients are a = -3, b = 150, c = -1200, hence:

• $\Delta = 150^2 - 4(-3)(-1200) = 8100$

• $x_1 = \frac{-150 + \sqrt{8100}}{-6}$

• $x_2 = \frac{-150 - \sqrt{8100}}{-6} = 40$

He needs to charge $40 to break even.

For a profit of $600, we have that P(x) = 600, hence:

-3x² + 150x - 1200 = 600.

-3x² + 150x - 1800 = 0.

The coefficients are a = -3, b = 150, c = -1800, hence:

• $\Delta = 150^2 - 4(-3)(-1800) = 900$

• $x_1 = \frac{-150 + \sqrt{900}}{-6}$

• $x_2 = \frac{-150 - \sqrt{900}}{-6} = 30$

He needs to charge $30 for a profit of $600.

More can be learned about quadratic equations at brainly.com/question/24737967

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