A restaurant has a fixed price of $36 for a complete dinner... Please help solving this problem :)
A restaurant has a fixed price of $36 for a complete
dinner. The average number of customers per night is
200. The owner estimates that for each dollar
increase in the price of the dinner, there will be, on
average, four fewer customers per night. Estimate the
price for the dinner that will produce the maximum
amount the owner can expect to receive
 A: Let $x$ represent the unit of change to price and customer expectancy.  
Define the revenue function based on the relationship $\textrm{revenue = price} \times \textrm{volume}$.  We have
$$R(x)=(36+x)(200-4x)$$
Since this is a quadratic function, we can find the maximum by converting the equation into vertex form.
\begin{align}
R(x)&=(36+x)(200-4x)\\
&=7200+56x-4x^2\\
&=-4(x^2-14x)+7200\\
&=-4(x^2-14x\color{blue}{+49-49})+7200\\
&=-4(x^2-14x+49)+196+7200\\
&=-4(x-7)^2+7396
\end{align}
Since the vertex to $R(x)$ is at $(7,7396)$, that means the maximum when $x=7$.  Substitute this into the price expression
$$36+x=36+(7)=43.$$
Therefore, the price that would produce the maximum revenue is $\$43$.
A: I think I messed up somewhere, but I'll give it a try:
The total money, given that the price increased by $n$ dollars, should be 
$$(36+n)(200-4n)=-4n^2+56n+7200.$$ 
Now, we have to maximize this. This makes sense to maximize because $a$ is negative. Now, for a general quadratic equation $ax^2+bx+c$ where $a<0$, the maximum occurs at 
$$x=-\frac b{2a}\text{ is: }c-\frac{b^2}{4a}$$,
$$ and you should be able to get your answer from here.
Hope this helps.
A: If the meal costs $36 + k$ dollars the owner will have $200 - 4k$ costumers.
So he will make $(36 + k)(200 - 4k)$ dollars.
What is the highest possible value of $(36+k)(200-4k)$?
.....
$(36+k)(200-4k) = 7200 + 56k - 4k^2=$
$7200 - (4k^2 -56k) = $
$7200 - ((2k)^2 - 2*2k*14) =$
$7200 + 32^2 - ((2k)^2 - 2*2k*32 + 14^2) =$
$7200 + 196 - (2k - 14)^2 =$
$7396 - (2k-14)^2 \le 8224$.
So if $(2k-14)^2 > 0$ we will have that $7396-(2k-14)^2 < 7396$.
If $(2k-32)^2 = 0$ we will have that $7396-(2k-14)^2 = 7396$.
If $(2k-14)^2 < 0$ we will have broken the laws of space and time and will all die horribly, alone, and in pain.
So the maximum profit revenue is $7396$ when $2k -14=0$.
SO if $k =7$ and he charges $36+7 = 43$ he will make the most money.
But he will lose $28$ costumers.  Hard luck for them I guess.....  All for just $196$.... Oh, $196$ per night is a significant amount of money.  To heck with those $28$ cheapskates...
