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.