This Is Confusing Me So Much. Help Please!?
the sweet drip beverage co., sells cans of soda pop in machines. it finds that sales average 26,000 cans per month when the cans sell for 50 cents each. for each nickel increase in the price, the sales per month drop by 1000 cans. Determine a function R(x) that models that total revenue realized by Sweet Drip, where x is the number of $0.05 increases in the price of a can. How much should Sweet Drip charge per can to realize the maximum revenue? What is the maximum revenue?
September 26th, 2009 at 3:15 am
without using calculus 26000-1000x cans sold per month $.50+$.05x amount collected for each can so:
total revenue =(26000-1000x)*(.50+.05x)
multiply that out(F.O.I.L. it) and you get:
revenue=13000+1300x-500x-50x^2
group like terms and relocate the terms:
-50x^2+800x+13000 (look familiar? its a Parabola formula)
then you just have to locate the vertex (because the vertex is always the min or the max)
the x value of the vertex is given by -b/2a so plugging in -50 for a and 800 for b:
-800/(2*-50)=8
so 8 is the x value of the vertex and you can plug it in to the equation to get y:
-50(8)^2+800(8)+13000=16200 for y
y is the maximum revenue and if you plug 8 into $.50+$.05x you get the price per can
$.50+$.05(8)=$.90
September 26th, 2009 at 7:08 am
Let S == sales (units /month)
Let R == revenue (assumed total sales, not profit)
Let P = price
S = linear function of price, descending as price increases
S = AP + B
26000 = A(50) + B
25000 = A(55) + B (sell 1000 less with 5cent higher price)
solve for A and B and you get
S = AP + B (not giving that part away)
R = SP (sales x price)
R = (AP+B) x P
maximize for R (easy with calculus, dunno if that is the class you are in)
September 26th, 2009 at 11:11 am
whhhhhuuuuuuuuuuuuu?