To solve this problem, we'll analyze the given function [tex]\( P(x) = -0.0005(x^2 + 50)(x-40)(x-100) \)[/tex] and determine the values of [tex]\( x \)[/tex] for which the profit is zero, and the range of [tex]\( x \)[/tex] for which the company makes a profit.
1. Finding the units where the profit is exactly [tex]$0$[/tex]:
The profit is exactly zero at the roots of the function [tex]\( P(x) \)[/tex], i.e., where [tex]\( P(x) = 0 \)[/tex]. The roots of the polynomial equation are found to be [tex]\( x = 40 \)[/tex] and [tex]\( x = 100 \)[/tex].
Therefore, the company's monthly profit will be exactly \[tex]$0 if it makes and sells 40 units or 100 units.
2. Determining the range where the company makes a profit:
The company makes a profit in the intervals where the profit function \( P(x) \) is positive. Since \( P(x) \) changes sign at its roots, the intervals need to be examined between and outside the roots.
By analyzing the graph or value changes between \( x = 40 \) and \( x = 100 \), we can determine that the company will make a profit if it makes and sells units between these two points.
Therefore:
1. The company's monthly profit will be exactly $[/tex]0[tex]$ if it makes and sells 40 or 100 units.
2. The company will make a profit if it makes and sells units in the interval (40, 100).
3. The company won't make a profit if it makes and sells units outside the interval (40, 100). Specifically:
- Less than 40 units
- More than 100 units
So, the correct answers are:
- The company's monthly profit will be exactly \$[/tex]0 if it makes and sells 40 or 100 units.
- The company will make a profit if it makes and sells between 40 and 100 units.
- The company won't make a profit if it makes and sells less than 40 units or more than 100 units.
