Maximizing Revenue

Question

Suppose the quantity demanded per week of a certain dress is related to the unit price p by the demand equation p=√(700-x), where p is in dollars and x is the number of dresses made. To maximize the revenue, how many dresses should be made and sold each week?

how many dresses?

Answer 1

Revenue = R = px = (sqr(700-x))x

R'(x) = sqr(700-x) - x/2sqr(700-x) = 0

sqr(700-x) = x/2sqr(700-x)

2(700-x) = x

1400-2x = x

3x =1400

x = 1400/3 = 466 2/3 dresses will maximize revenue

max Revenue = sqr(700-466 2/3)(466 2/3) = slightly more than $7128.45

for 500 dresses R=sqr200)(200) = $2828,43

for 400 dresses R= sqr300(300) = $5196.15

for 466 dresses R =sqr234(466) = $7128.43

for 467 dresses R = sqr233(467) = $7128.445