The profit made when t units are sold, t>, is given by P²-36t+320. Determine the number of units to be sold in order for P>0 (a profit is made).
I do not understand your P^2 - 36t+320.
where is the profit equation?
Im sorry i typed it wrong.
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).
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 that satisfy this condition.
Given the profit equation P² - 36t + 320, we can rewrite it as P² - 36t + 320 > 0.
To solve this inequality, we can factorize the quadratic equation if possible. However, in this case, the quadratic equation cannot be factored easily.
Another approach is to use the quadratic formula. The quadratic formula can be expressed as:
t = (-b ± √(b² - 4ac)) / (2a)
In our case, the equation is P² - 36t + 320 = 0, so a = 1, b = -36, and c = 320.
Substituting these values into the quadratic formula, we get:
t = (-(-36) ± √((-36)² - 4(1)(320))) / (2(1))
t = (36 ± √(1296 - 1280)) / 2
t = (36 ± √16) / 2
t = (36 ± 4) / 2
This gives us two possible solutions for t:
1. t = (36 + 4) / 2 = 40 / 2 = 20
2. t = (36 - 4) / 2 = 32 / 2 = 16
So, there are two potential values of t: 20 and 16, which correspond to the number of units to be sold in order for a profit to be made (P > 0).