Chapter 9.6, Problem 50STP

To determine

To find: the time taken by tank to drain out.

Answer to Problem 50STP

  $t\approx 4.586$ minutes

Explanation of Solution

Given:

  $\begin{array}{l}\text{Amount of water in tank is }L=-4t^2-10t+130\\ L=\text{Number of liters of water}\end{array}$

Formula Used:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{4a}$

Calculation:

  $\begin{array}{l}L\left(t\right)=-4t^2-10t+130\\ L\left(0\right)=-{4.0}^2-10\times 0+130=130\\ \\ L\left(t\right)=0\\ -4t^2-10t+130=0\\ -2\left(2t^2+5t-65\right)=0\\ 2t^2+5t-65=0\\ t\frac{-5\pm \sqrt{5^2-4\times 2\times \left(-65\right)}}{2\times 2}=\frac{-5\pm \sqrt{545}}{4}\\ =\frac{-5\pm 23.345}{4}\\ t_1=\frac{-5-23.345}{4}\approx -7.086\\ t_2=\frac{-5-23.345}{4}\approx 4.586\\ t\approx 4.586\end{array}$

Conclusion:

Thus the time is $t\approx 4.586$