High School

The population [tex]P(t)[/tex] of a bacteria culture is given by [tex]P(t) = -1500 t^2 + 60000 t + 10000[/tex], where [tex]t[/tex] is the time in hours after the culture is started.

Determine the times at which the population will be greater than 460,000 organisms.

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.