The weekly revenue of a business selling gummy bear bags is a function of price.

Question

The weekly revenue of a business selling gummy bear bags is a function of price. Weekly revenue is $8,208 when the price is $1.90 but is $2,800 when the price is $3.50 Find a quadratic model that fits this information.Also know that the revenue is $0 if the price is $0.Let  = the price of a bag of gummy bears. Then:

Revenue = = ______ 

Answer 1

(2,82), (3.5, 28), (0,0) are close to points on the graph, just to simplify the number

with the y coordinate in hundreds, x coordinate in dollars

0=ah^2 + k

28 =a(3.5-h)^2 + k

82 =a(2-h)^2 + k

3 equations, 3 unknowns, solve by substitution and elimination or by matrix algebra

28-=12.25a - 7ah + ah^2 + k

82 = 4a -4ah+ ah^2 + k subtract

54 = -8.25a +3ah

subtract 1st equation 0=ah^2 + k from 28=12.25a -7ah+ah^2 + k

28= 12.25a -7ah multiply by 3

84 = 36.75a -21ah

multiply 54=-8.25a +3ah by 7 and add to 84=36.75a-21ah

378 = -57.75a + 21ah

462 = -21a

a =462/-21 = -22

R=-22(x-h)^2 + k

ah^2 + k =0

k= -ah^2

28 = -22(2-h)^2 + k = -22(4-4h+h^2) +22h^2 = -88+88h -22h^2 +22h^2 = -88 + 88h

88h = 116

h = 116/88 = 58/44 = 29/22 = about 1.3

k =22(29/22) = 29

vertex of the quadratic is (1.3, 29)

quadratic is roughly R(P) = -22(P-1.3,)^2 + 29

or in thousands and dollars

R(P) = -22(P-$1.30)^2 + 2900

at a price of $1.30, revenue is maximized at $2,900

this is a little off, given that we used 2 instead of 1.9, and 3.5 instead of 3.7 and 8200 instead of 8202, but you know the vertex (h,k) has 0<h<3.7 and k >2800. h=1.3 and k= 2900 is in the right range. You also know a<0 as it's a downward opening parabola with vertex as the maximum point. absolute value of a is fairly large, as the parabola is narrow.

work it the same way with the actual numbers to get a more accurate quadratic equation

General equation is R(P) = a(P-h)^2 + k where (h,k) is the vertex