Graph the inequalities given by the set of constraints. Find points where the boundary lines intersect to form a polygon. Substitute the coordinates of each point into the objective function and find the one that results in the largest value.

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Noah has a summer tree-trimming business. Based on experience, Noah knows that his profit, P, in dollars, can be modelled by = −

crimeas [40]

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.

<h3>What is a quadratic function?</h3>

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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