The environmental club sells sweatshirts as a fund raiser . They sell 1200 shirts a year at $ 20 each. They are planning to increase the price.A suevey indicates that , for every $ 2 increase in price, there will be a drop of 60 sales a year. What should the selling price be in order to maximize the revenue.
2006-12-07 03:04:50 · 5 answers · asked by SHAILY 1 in Science & Mathematics ➔ Mathematics
The new price will be (20 + 2x).
The new sales volume will be (1200 - 60(2x)).
The new profit will be:
P(x) = (20 + 2x)(1200 - 120x)
P(x) = 2(10 + x)120(10 - x)
P(x) = 240(10 + x)(10 - x)
P(x) = 240(100 +10x - x²)
To maximize the function, you can take the derivative and set it equal to zero:
P'(x) = 10 - 2x = 0, so x = 5 or $5.00
So the new price for maximum profit would be $20 + 2($5) = $30
If you don't understand derivatives, you can put it in the form of a parabola, and the vertex is the maximum profit.
2006-12-07 03:14:42 · answer #1 · answered by Jim Burnell 6 · 0⤊ 0⤋
The maximum revenue is made by selling 900 shirts at $30 each. Total revenue will be $27000.
2006-12-07 11:12:49 · answer #2 · answered by MrBill 2 · 0⤊ 0⤋
r(n) = (selling price)*(number of shirts)
r(n) = (20+2n) * (1200-60n) = -120(n^2)+1200n+24000
r'(n) = -240n+1200
r'(n) = 0 ==> n=1200/240 =5
selling price = 20+2*5 =30
2006-12-07 11:21:27 · answer #3 · answered by Hamidreza 2 · 0⤊ 0⤋
900 shirts at $30 a piece.
2006-12-07 11:11:01 · answer #4 · answered by JP 2 · 0⤊ 0⤋
the answer is 800.
dude just guessing it out
2006-12-07 11:11:14 · answer #5 · answered by Aditya N 2 · 0⤊ 0⤋