1. **State the problem:** We are given the quadratic function $$p(t) = -1.875 t^2 + 30 t - 20$$ which represents profit as a function of time $$t$$. 2. **Identify the type of function:** This is a quadratic function of the form $$p(t) = at^2 + bt + c$$ where $$a = -1.875$$, $$b = 30$$, and $$c = -20$$. 3. **Important rule:** Since $$a < 0$$, the parabola opens downward, so the vertex represents the maximum profit. 4. **Find the vertex:** The time $$t$$ at which the maximum profit occurs is given by the formula $$t = -\frac{b}{2a}$$. 5. **Calculate $$t$$:** $$t = -\frac{30}{2 \times (-1.875)} = -\frac{30}{-3.75} = \cancel{\frac{30}{3.75}} = 8$$ 6. **Find the maximum profit $$p(8)$$:** $$p(8) = -1.875 \times 8^2 + 30 \times 8 - 20 = -1.875 \times 64 + 240 - 20 = -120 + 240 - 20 = 100$$ 7. **Interpretation:** The maximum profit is 100 units at time $$t = 8$$. 8. **Domain note:** Since $$t$$ is in the interval $$[0,15]$$, the vertex at $$t=8$$ lies within this domain, confirming the maximum profit occurs within the given time frame.