# 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?

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Most people the world over probably don't know the abbreviation "Dhs" for darahim. –  joriki Feb 27 '13 at 17:57
dhs= darahim ____ thanks joriki –  moamen Feb 27 '13 at 18:00
can you help me to answer this question?? –  moamen Feb 27 '13 at 18:01
The currency involved doesn't matter. Whatever it is, we want more of it. –  Ross Millikan Feb 27 '13 at 18:43

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.

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nice >>> thanks alex --- now i reading your answer –  moamen Feb 27 '13 at 18:40
Hints: profit = total revenue - total cost, total revenue = (number of dozens sold)$\cdot$(price per dozen), and total cost = (number of dozens sold)$\cdot$(cost per dozen). The variable $x$ represents the number of dozens sold. Find a function for profit as a function of $x$, then maximize using calculus (if this is a calculus class) or a calculator (if it isn't).
Graph the function $y=P(x)$, then use your calculator's "maximum" function (it will be different on every calculator - for TI83/84/86 it will be in the "calc" menu). –  icurays1 Feb 27 '13 at 18:31