Question

Application of Quadratic Equations

Business The manager of a bicycle shop has found that, at a price (in dollars) of p(x)=150-x/5 per bicycle, x bicycles will be sold.

a)Find an expression for the total revenue from the sale of x bicycles. (Hint: Revenue = Demand X Price.)
b)Find the number of bicycle sales that leads to maximum revenue

Answer 1

Price function is \[P(x) = 150-x/5\] or \[P(x) = \frac{150-x}{5}\]?

Answer 2

how can i give you a medal

Answer 3

oh ok.so should i post the second question and tag u to it

Answer 4

this is the question, is it not?

Answer 5

yes it is

Answer 6

okay. so, I asked for clarification about the form of the price function. which one is it?

Answer 7

I'm guessing perhaps the first one, because the second could be written as 30-x/5, but I've found that people tend not to heed the operator precedence when copying problems to the forum, and it could really be either one.

Answer 8

dont reallu=y get you clearly.cos i got ma a) as x(150-x/5)=150x -x^2/5

Answer 9

Revenue = number bikes sold (x) * price of bike (150-x/5) = 150x - x^2/5

now we need to find the number of bikes sold that maximizes revenue. Again, this is a parabola...

Answer 10

yeah i thought as much

Answer 11

ok with vertex h=-b/2a

Answer 12

h=x^2/5 /2(150) and this is where i got confused

Answer 13

remember, a and b are the coefficients, and don't include the variable itself.

Answer 14

\[y = ax^2+bx+c\]vertex at \((-b/2a, y(-b/2a))\)

\[y = -\frac{1}{5}x^2+150x\]\[a=-\frac{1}{5}, b = 150\]
\[-\frac{150}{2*(-\frac{1}{5})} = 150 * \frac{5}{2} = \]

Answer 15

oh ok. so if im to find the maximum revenue i would need to put 375 in the very first revenue equation we got right? thus150(375)-(375)^2/5

Answer 16

Yes, that should do the trick.

Answer 17

perfect.