# Thread: Last Quadratic work problem

1. ## Last Quadratic work problem

a hole is punched in the side of a cylindrical water tank and water begins to flow out. The hole is punched at time t=0 min. The volume of water in the tank, V litres, at any time t when the water is flowing out is given by V=t^2 - 16t + 100 .

a. how much water is in the tank at the beginning? i got 100 litres bt from my calculations from the next question I think it may b wrong.
b. How long does it take to empty the tank?is this just completing the square?
c. How much water flows out of the tank and how much is left inside?
d. Determine a formula for the time t when the tank has a given volume V. State the domain and range of this function.

2. 1. Yep, that's right

2. Set $t^2-16t+100=0$
If you check the discriminant of the expression you get $\Delta = (-16)^2-4 \times 100 \times 1= -144 < 0$, so there are no real solutions, and hence V never reaches 0. The tank never empties completely.

3. You want to find the lowest value of V. Therefore, you need to find the turning point of the graph $V=t^2-16t+100$.
Just complete the square here:

$V = t^2-16t+64-64+100$

$V = (t-8)^2+36$

Hence, the turning point is located at (8, 36). So 64 litres flow out of the tank and 36 litres remains.

4. Simply solve for t. Fortunately having it in 'complete the square' form makes it a lot easier to solve.

$(t-8)^2=V-36$

$t-8=\pm \sqrt{V-36}$

$t=8 \pm \sqrt{V-36}$

Notice since the parabola curves upwards again you will get two values of t for the one value of V.