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Answer :
To determine the times at which the population of the bacteria culture will be greater than 460,000 organisms, we start with the given population function:
[tex]\[ P(t) = -1500t^2 + 60000t + 10000 \][/tex]
We need to find the values of [tex]\( t \)[/tex] where:
[tex]\[ P(t) > 460,000 \][/tex]
Substitute 460,000 into the inequality:
[tex]\[ -1500t^2 + 60000t + 10000 > 460,000 \][/tex]
Subtract 460,000 from both sides to form a quadratic inequality:
[tex]\[ -1500t^2 + 60000t + 10000 - 460,000 > 0 \][/tex]
Simplify the left side:
[tex]\[ -1500t^2 + 60000t - 450,000 > 0 \][/tex]
Now, we need to solve for [tex]\( t \)[/tex] where this inequality holds true. Solving the corresponding quadratic equation helps us identify intervals:
[tex]\[ -1500t^2 + 60000t - 450,000 = 0 \][/tex]
Dividing the equation by -1500 for simplification results in:
[tex]\[ t^2 - 40t + 300 = 0 \][/tex]
Next, solve for the roots using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -40 \)[/tex], and [tex]\( c = 300 \)[/tex].
[tex]\[ t = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(1)(300)}}{2(1)} \][/tex]
[tex]\[ t = \frac{40 \pm \sqrt{1600 - 1200}}{2} \][/tex]
[tex]\[ t = \frac{40 \pm \sqrt{400}}{2} \][/tex]
[tex]\[ t = \frac{40 \pm 20}{2} \][/tex]
This gives solutions:
[tex]\[ t = \frac{60}{2} = 30 \quad \text{and} \quad t = \frac{20}{2} = 10 \][/tex]
The solutions to the quadratic inequality indicate that the population will be greater than 460,000 organisms in the interval between these roots. Therefore:
[tex]\[ 10 < t < 30 \][/tex]
This means the times at which the bacteria population will be greater than 460,000 organisms are between 10 and 30 hours.
[tex]\[ P(t) = -1500t^2 + 60000t + 10000 \][/tex]
We need to find the values of [tex]\( t \)[/tex] where:
[tex]\[ P(t) > 460,000 \][/tex]
Substitute 460,000 into the inequality:
[tex]\[ -1500t^2 + 60000t + 10000 > 460,000 \][/tex]
Subtract 460,000 from both sides to form a quadratic inequality:
[tex]\[ -1500t^2 + 60000t + 10000 - 460,000 > 0 \][/tex]
Simplify the left side:
[tex]\[ -1500t^2 + 60000t - 450,000 > 0 \][/tex]
Now, we need to solve for [tex]\( t \)[/tex] where this inequality holds true. Solving the corresponding quadratic equation helps us identify intervals:
[tex]\[ -1500t^2 + 60000t - 450,000 = 0 \][/tex]
Dividing the equation by -1500 for simplification results in:
[tex]\[ t^2 - 40t + 300 = 0 \][/tex]
Next, solve for the roots using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -40 \)[/tex], and [tex]\( c = 300 \)[/tex].
[tex]\[ t = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(1)(300)}}{2(1)} \][/tex]
[tex]\[ t = \frac{40 \pm \sqrt{1600 - 1200}}{2} \][/tex]
[tex]\[ t = \frac{40 \pm \sqrt{400}}{2} \][/tex]
[tex]\[ t = \frac{40 \pm 20}{2} \][/tex]
This gives solutions:
[tex]\[ t = \frac{60}{2} = 30 \quad \text{and} \quad t = \frac{20}{2} = 10 \][/tex]
The solutions to the quadratic inequality indicate that the population will be greater than 460,000 organisms in the interval between these roots. Therefore:
[tex]\[ 10 < t < 30 \][/tex]
This means the times at which the bacteria population will be greater than 460,000 organisms are between 10 and 30 hours.
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