# Maximum of d(12-d)

I'm a little confused on a quite simple quadratic problem. I need to calculate the maximum of $d(12-d)$ using basic quadratics. The answer is $6$ as can also be shown by $f'(x)= -2d +12$, however this is algebra not calculus. Using the squares method I get twelve, I think I may be using the incorrect method. Could someone set me on the right track?

$d(12-d)$
$= -d^2 + 12d$
$-d^2 + 12/(-2)d + 36 = 36$
$(d-6)^2 = 36$
$d = 12 or 0$ <- I understand these are the roots, am I not meant to calculate them?

The suggested answer is: $-(d-6)^2 + 36$, but I have no idea where it came from.

$$d(12-d)=12d-d^2=-(d^2-2\cdot6\cdot d+6^2)+6^2=36-(d-6)^2$$

Now for real $d, (d-6)^2\ge 0$

$\implies d(12-d)\le 36$

• How can I use this to get the maximum value of 6? Thanks
– Sam
May 18, 2013 at 6:47
• @Sam, we need to maximize $6(12-d),$ a constant can not be changed, hence can not be maximized. Now, $(d-6)^2\ge 0\implies -(d-6)^2\le 0\implies 36-(d-6)^2\le 36$ May 18, 2013 at 6:51

You can use your knowledge of parabolas. The parabola $y=x(12-x)$ opens downwards, and reaches its maximum halfway between the two points where it crosses the $x$-axis. These two points can be read off from the equation.

• Do you think this is how the suggested answer was obtained?
– Sam
May 18, 2013 at 6:30
• It may have been: If I wrote down an expression like it, such as $(d-7)(17-d)$, I would know the max was at $12$. But you might be expected to solve the problem by completing the square, as in the answer by lab bhattacharjee. In schools, well before calculus, the parabola approach might be expected. But since you attempted a completing the squares approach, I would lean to that one. May 18, 2013 at 6:38