To raise money, students sell candy grams.

Question

ENTIRE QUESTION:

To raise money, students sell candy grams. From past experience, they expect to sell 400 candy grams at a price of $4. They have also learned from experience that each $0.50 increase in price causes a drop in sales of 20 candy grams

a.Write an equation that models the situation

.b.Suppose the student council needs a revenue of at least $1800. Write and solve the inequality to find all the possible prices that will achieve this goal.

c.Show how your solution would change if the student council needed to raise $1600 or more.

Answer 1

(400,4) is a point on the demand curve (straight line in this example)

slope = m = .5/-20 = -5/200= -1/40

y=mx +b or P=-x/40 + b where P = price, x =sales and b is calculated by plugging in the point (400,4) 4=-400/40+b, b=4+10=-$14. At $14, no one buys anything.

P=-x/40 + 14 is the equation of the demand curve

Revenue = R= Px = -x^2/40 +14x

1800< -x^2/40 +14x

x^2/40 -14x + 1800 < 0

x^2 - 560x + 72000 < 0

x<560/2 + or - (1/2)sqr(560^2 -4(72,000)) = 280+80= 200 or 360

P=-200/40+14 = 14-5 = $9

or -360/40 +14 =14-9= $5

they need to set price at between $5 and $9 to get revenue of at least $1,800

R=-x^2/40 +14x

R'=-x/20 +14 = 0

x=280

R=-280^2/40 +14(280) = 3920- 1960=$1960 = max revenue possible

P=-x/40+14 = -280/40 +14 = 14-7 = $7. The $5 and $9 prices are equidistant to $7

For revenue of at least $1600

x^2/40 -14x + 1600 < 0

x^2 -560x + 64000 < 0

x =560/2+ or - (1/2)sqr(560^2-4(64000) = 280+ or - 120 = 160 or 400

P=-(160)/40+14 = 14-4 = $10

or -400/40 +14 =14-10 =$4

To raise $1600, price needs to be between $4 and $10 (which are equidistant from $7 the revenue maximizing price)