Chapter 1.5, Problem 116E

Fish Population A large pond is stocked with fish. The fish population P is modeled by the formula $P=3t+10\sqrt{t}+140$, where t is the number of days since the fish were first introduced into the pond. How many days will it take for the fish population to reach 500?

To determine

The number of days it will take for the fish population to reach $500$.

Answer to Problem 116E

The number of days it will take for the fish population to reach $500$ is $\underset{\_}{89\text{days}}$.

Explanation of Solution

Given:

The fish population in the tank is modeled by the formula $P=3t+10\sqrt{t}+140$, where, $P$ is number of fish and t is the number of days since the fish were first introduced into the pond.

Formula used:

Quadratic formula:

The solution of a quadratic equation of the form $a{x}^2+bx+c=0,a\ne 0$ can be obtained by using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

Substitute $500$ for $P$ in $P=3t+10\sqrt{t}+140$ to obtain the number of days it will take for the fish population to reach $500$.

$\begin{array}{l}500=3t+10\sqrt{t}+140\\ 3t+10\sqrt{t}+140-500=0\\ 3t+10\sqrt{t}-360=0\\ 3{\left(\sqrt{t}\right)}^2+10\sqrt{t}-360=0\end{array}$

Let $x=\sqrt{t}$. Substitute $x$ for $\sqrt{t}$ in $3{\left(\sqrt{t}\right)}^2+10\sqrt{t}-360=0$.

$\begin{array}{l}3{\left(x\right)}^2+10x-360=0\\ 3x^2+10x-360=0\end{array}$

Use Quadratic formula to find the value of x.

Substitute $3$ for $a$, $10$ for $b$ and $-360$ for $c$ in $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

$\begin{array}{c}x=\frac{-\left(10\right)\pm \sqrt{{\left(10\right)}^2-4\times 3\times \left(-360\right)}}{2\times 3}\\ =\frac{-10\pm \sqrt{100+4320}}{6}\\ =\frac{-10\pm \sqrt{4420}}{6}\end{array}$

Simplify the above equation as follows.

$\begin{array}{c}x=\frac{-10\pm 66.48}{6}\\ =\frac{-10+66.48}{6}\text{or }\frac{-10-66.48}{6}\\ =\frac{56.48}{6}\text{or }\frac{-76.48}{6}\\ =9.41\text{or }-12.75\end{array}$

Note that, the number of days must be positive.

Therefore, substitute $\sqrt{t}$ for $x$ in $x=9.41$.

$\begin{array}{l}\sqrt{t}=9.41\\ t={\left(9.41\right)}^2\\ t=88.5481\\ t\approx 89\end{array}$

Thus, the number of days it will take for the fish population to reach $500$ is $\underset{\_}{89\text{days}}$.