The profit made when t units are sold, t>0, is given P=t²-36t+320. Determine the number of units to be sold in order for P>0 (a profit is made).
Try graphing the equation and figuring out t from there.
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To determine the number of units to be sold in order for a profit to be made (P > 0), we need to find the values of t for which the given profit function is greater than zero.
The profit function P is defined by: P = t² - 36t + 320
To find when P > 0, we can solve the quadratic inequality:
t² - 36t + 320 > 0
To solve this inequality, we can first find the roots of the equation t² - 36t + 320 = 0, and then identify the intervals where P > 0.
To find the roots of the equation t² - 36t + 320 = 0, we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
In this case, a = 1, b = -36, and c = 320. Substituting these values into the quadratic formula:
t = (-(-36) ± √((-36)² - 4(1)(320))) / (2(1))
Simplifying:
t = (36 ± √(1296 - 1280)) / 2
t = (36 ± √16) / 2
t = (36 ± 4) / 2
This gives us two possible solutions:
t1 = (36 + 4) / 2 = 40 / 2 = 20
t2 = (36 - 4) / 2 = 32 / 2 = 16
These are the roots of the equation t² - 36t + 320 = 0.
Now, let's consider the intervals:
When t < t1 = 20, the profit function P is negative.
When t > t2 = 16, the profit function P is positive.
Therefore, for the profit function to be greater than zero (P > 0), the number of units to be sold must be t > 16.