# Math Help - [SOLVED] Completing the Square Problem?

1. ## [SOLVED] Completing the Square Problem?

"In an electrical circuit, the voltage, V volts, as a function of time, t minutes, is given by $V = 12 - 9t + 2t^2$. Determine the greatest and least values of voltage during the first 5 min. When do these values occur?"

The first thing I did was substitute $t = 5$.

$
V = 12 - 9t + 2t^2$

$V = 12 - 9(5) + 2(5)^2$
$V = 12 - 45 + 50$
$V = 17$

Then I go check the answer page to see if I was right. The answer is "Least value: 1.875 V at 2.25 min; greatest value: 17 V at 5 min". How was the first part of the answer obtained? Is completing the square involved? If it is, I already have that done.

$
V = 12 - 9t + 2t^2$

$V - 12 - 2 = 2(t^2 - \frac{9}{2}t)$
$V - 12 - 2(\frac{9}{4}) = 2(t^2 - \frac{9}{2}t + \frac{9}{4})$
$V - 12 - \frac{18}{4} = 2(t - \frac{3}{2})^2$
$V - \frac{48}{4} - \frac{18}{4} = 2(t - \frac{3}{2})^2$
$V - \frac{66}{4} = 2(t - \frac{3}{2})^2$
$V = 2(t - \frac{3}{2})^2 + \frac{33}{2}$

I'm guessing that this wasn't needed, as this seems to have no relevance to the aforementioned answer. What needs to be done here? Help would be appreciated.

2. You made a couple of errors when completing the square, and this is actually a perfectly valid method of solving the problem.

$V - 12 = 2(t^2 - \frac{9}{2}t)$

$V - 12 + 2 \cdot \frac{81}{16} = 2(t^2 - \frac{9}{2}t + \frac{81}{16})$

$V - 12 + \frac{81}{8} = 2(t - \frac{9}{4})^2$

$V - \frac{96}{8} + \frac{81}{8} = 2(t - \frac{9}{4})^2$

$V - \frac{15}{8} = 2(t - \frac{9}{4})^2$

$V = 2(t - \frac{9}{4})^2 + \frac{15}{8}$

and the minimum value is obtained when $t = \frac{9}{4}$,
from which the desired conclusion follows.

3. Ah, got it now. I forgot to square 9/4. Thanks very much for the help.