Question 965267
PRICE*UNITSSOLD=REVENUE


15*6000=R
(15-1)*(6000+1000)=R(1)
(15-2)(6000+2*1000)=R(2)
(15-3)(6000+3*1000)=R(3)
.
.
The variable here is the amount of decrease from the initial 15 dollar per gallon price, so the variable choice this way can be x, the amount of price DECREASED off the $15.


Formula for revenue is the function {{{R(x)=(15-x)(6000+1000x)}}}.


Do the simplifying into general form, and you'll have the quadratic equation to which you can apply the general solution for a quadratic equation; but what you want is the value for x in the MIDDLE of the zeros of R.  That will be the maximum, because R(x) is a function with a maximum point  (according to the coefficient on x^2 being negative).


If that analysis makes sense, after you read and think about it carefully, then you know what to do.  Just remember, as analyzed here, the price you are looking for is 15 MINUS the middle value between the zeros.


Note, {{{R(x)=1000(15-x)(6+x)}}}, so when you examine the roots, 
{{{0=1000(15-x)(x+6)}}}
Dividing both sides by 1000,
{{{(x+6)(15-x)=0}}}


...and again, look for the x value exactly between the two zeros.