P=t^2-35t+300 the profit made when t units are sold t>0, is given by
P=t^2-35t+300. Determine the number units to be sold in order for P<0 (a loss is taken).
How do I solve this? Thanks for your help.
well, when is t^2 -35 t + 300 = 0
plain old quadratic
t = [ 35 + / - sqrt (1225 - 1200) ]/2
= [ 35 +/- 5 ]/2
= 15 or 20
your profit is negative between 15 units and 20 units
I suspect a typo in your problem. It does not make sense to have a profit of 300 for zero units sold.
To solve the inequality P < 0, we need to find the values of t for which the profit is negative, indicating a loss.
The given equation for profit is P = t^2 - 35t + 300. In order to determine the number of units to be sold when a loss is taken, we need to find the values of t that make P less than zero.
To do this, we can:
1. Set P = 0:
0 = t^2 - 35t + 300
2. Factorize the equation:
0 = (t - 25)(t - 10)
3. Set each factor equal to zero:
t - 25 = 0 or t - 10 = 0
4. Solve for t:
t = 25 or t = 10
We have found that the profit is zero at t = 25 and t = 10.
To determine when the profit is negative (P < 0), we need to evaluate the profit equation for values of t between t = 10 and t = 25.
For example, when t = 15:
P = (15)^2 - 35(15) + 300
= 225 - 525 + 300
= 0
We can see that the profit is zero when t = 15, which means it is neither negative nor positive.
To find the values of t for which P < 0, we need to consider the range of t's between t = 0 and t = 10, as well as the range between t = 25 and beyond.
Therefore, to have a negative profit (a loss), the number of units to be sold must be less than 10 or greater than 25.