jenny farm and the dozen egg ??? Farmer Jenny decides to expand her business interests and starts to package and sell the eggs produced by her chooks to a local shop.
The cost of producing $x$ dozen eggs per day is given by, in dirhams:
$$C=( 5/12 x^2+ 4x-3)$$ 
and the selling price of one dozen eggs is given by, in dirhams$$(6- 1/4 x)$$
How many dozen eggs should be produced each day to maximize the total profit?

Expanding on the answer by icurays1, the goal is indeed to maximize the quadratic, and the easiest way to do this uses calculus.
But in case you haven't learned calculus, you can maximize a quadratic just by completing the square - no calculator necessary!
The profit for selling $x$ dozen eggs is given by 
\begin{align}
p(x) &= x\cdot \text{(price per dozen)} - \text{cost to make $x$ dozen}\\
&= x(6-\frac{1}{4}x) - (\frac{5}{12}x^2+4x-3)\\
&= -\frac{2}{3}x^2 + 2x + 3\\
&= -\frac{2}{3}(x^2 - 3x) + 3\\
&= -\frac{2}{3}(x-\frac{3}{2})^2 + \frac{3}{2} + 3\\
&= -\frac{2}{3}(x-\frac{3}{2})^2 + \frac{9}{2}
\end{align}
Now the term $-\frac{2}{3}(x-\frac{3}{2})^2$ is always negative or zero, so the quadratic achieves its maximum value when $-\frac{2}{3}(x-\frac{3}{2})^2 = 0$, i.e. when $x = \frac{3}{2}$.
Farmer Jenny should produce $\frac{3}{2}$ dozen eggs each day.
The trick here is to get rid of the linear term by "completing the square" so that you can easily see where the maximum is.
