To find the time [tex]\( t \)[/tex] at which the beekeeper should harvest the honey to maximize profit, we need to analyze the quadratic equation:
[tex]\[ P(t) = -16t^2 + 2050t + 150 \][/tex]
This is a standard quadratic equation of the form [tex]\( P(t) = at^2 + bt + c \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 2050 \)[/tex], and [tex]\( c = 150 \)[/tex].
### Step-by-Step Solution
1. Identify the Coefficients:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 2050 \)[/tex]
- [tex]\( c = 150 \)[/tex]
2. Determine the Formula for the Vertex:
- The maximum (or minimum) of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] occurs at [tex]\( t = -\frac{b}{2a} \)[/tex].
3. Plug the Coefficients into the Vertex Formula:
- Substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ t = -\frac{2050}{2 \cdot (-16)} \][/tex]
- Calculate the denominator:
[tex]\[ 2 \cdot (-16) = -32 \][/tex]
- Divide the numerator by the denominator:
[tex]\[ t = -\frac{2050}{-32} \][/tex]
- Simplify the fraction:
[tex]\[ t \approx 64.0625 \][/tex]
4. Maximum Profit:
- To find the maximum profit, substitute [tex]\( t = 64.0625 \)[/tex] back into the original equation [tex]\( P(t) \)[/tex]:
[tex]\[ P(64.0625) = -16(64.0625)^2 + 2050(64.0625) + 150 \][/tex]
- Evaluate each term:
- [tex]\( (64.0625)^2 \approx 4103.89 \)[/tex]
- [tex]\( -16 \times 4103.89 \approx -65662.24 \)[/tex]
- [tex]\( 2050 \times 64.0625 \approx 131728.13 \)[/tex]
- [tex]\( -65662.24 + 131728.13 + 150 \approx 65814.0625 \)[/tex]
Hence, the maximum profit is approximately 65814.0625 dollars.
### Conclusion
The beekeeper should wait approximately 64.0625 days to harvest the honey, in order to achieve a maximum profit of approximately 65814.0625 dollars.
