SOLUTION: Each year, a baseball team sells boxes of chocolates as a fundraiser to lower the cost of team fees. The price of the chocolates and the number of boxes sold varies each year. The

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Question 1178346: Each year, a baseball team sells boxes of chocolates as a fundraiser to lower the cost of team fees. The price of the chocolates and the number of boxes sold varies each year. The information from five years of sales is given in the table below.
Price per box ($)= 3.00/4.00/5.50/6.50/8.00
Boxes sold= 5303/3360/1777/1316/1019
a. Find the regression equation in the form y=ax^3+bx^2+cx+d that best approximates the data. Express the values of a, b, c, and d to the nearest hundredth.
b. Use the equation to find the number of boxes, to the nearest whole number, that the team will sell if they charge $4.35 per box.
c. One year, the team only sold 125 boxes of chocolates. What price did they charge for each box?
d. If the team raises the price too high, they will not sell any boxes. Use your regression equation to predict the price of a box that will result in zero boxes sold.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

a. Find the regression equation in the form

y=ax%5E3%2Bbx%5E2%2Bcx%2Bd........Price per box ($)= 3.00 and Boxes sold= 5303
5303=a%2A3%5E3%2Bb%2A3%5E2%2Bc%2A3%2Bd
5303=27a%2B9b%2B3c%2Bd...........solve for c
3c=5303-27a-9b-d
c=5303%2F3-9a-3b-d%2F3...........eq.1


y=ax%5E3%2Bbx%5E2%2Bcx%2Bd........Price per box ($)= 4.00 and Boxes sold= 3360
3360=a%2A4%5E3%2Bb%2A4%5E2%2Bc%2A4%2Bd
3360=64a%2B16b%2B4c%2Bd...........solve for c
4c=3360-64a-16b-d
c=840-16a-4b-d%2F4...........eq.2


y=ax%5E3%2Bbx%5E2%2Bcx%2Bd........Price per box ($)= 5.50 and Boxes sold= 1777
1777=a%2A5.5%5E3%2Bb%2A5.5%5E2%2Bc%2A5.5%2Bd
1777=166.375a%2B30.25b%2B5.5c%2Bd...........solve for c
5.5c=1777-166.375a-30.25b-d
c=1777%2F5.5-166.375a%2F5.5-30.25b%2F5.5-d%2F5.5
c=323.09-30.25a-5.5b-d%2F5.5...........eq.3


y=ax%5E3%2Bbx%5E2%2Bcx%2Bd........Price per box ($)= 6.50+and Boxes sold= 1316
1316=a%2A6.5%5E3%2Bb%2A6.5%5E2%2B6.5c%2Bd........solve for c
6.5c=1316-a%2A6.5%5E3-b%2A6.5%5E2-d
c=1316%2F6.5-a%2A6.5%5E3%2F6.5-b%2A6.5%5E2%2F6.5-d%2F6.5
+c=202.462-a%2A6.5%5E2-6.5b-d%2F6.5........eq.4



from eq.1 and eq.2
5303%2F3-9a-3b-d%2F3=840-16a-4b-d%2F4..........solve for d
d%2F4-d%2F3=840-16a-4b-%285303%2F3-9a-3b%29
-d%2F12=840-16a-4b-5303%2F3%2B9a%2B3b
-d%2F12=+-7a+-+b+-+2783%2F3
-d%2F12=+-%287a+%2B+b+%2B+2783%2F3%29
d=+12%287a+%2B+b+%2B+2783%2F3%29
d=+84a+%2B+12b+%2B+11132....................eq.5

go to
c=5303%2F3-9a-3b-d%2F3...........eq.1, substitute +d
c=5303%2F3-9a-3b-%2884a+%2B+12b+%2B+11132%29%2F3
c=-37a+-+7+b+-+1943........................1)
go to
c=323.09-30.25a-5.5b-d%2F5.5...........eq.3, substitute d
c=323.09-30.25a-5.5b-%2884a+%2B+12b+%2B+11132%29%2F5.5
c=-45.5227+a+-+7.68182+b+-+1700.91............2)


from eq.1) and 2)
-37a+-+7+b+-+1943=-45.5227+a+-+7.68182+b+-+1700.91........solve for b
b+=+355.064+-+12.4999a...............3)
c=-37a+-+7+b+-+1943........................1) substitute +b
c=-37a+-+7+%28355.064+-+12.4999a%29+-+1943
c=50.4993+a+-+4428.45...................................4)

go to
d=+84a+%2B+12b+%2B+11132....................eq.5,substitute +b
d=+84a+%2B+12%28355.064+-+12.4999a%29%2B+11132
d=15392.8+-+65.9988a........................5)
go to
c=202.462-a%2A6.5%5E2-6.5b-d%2F6.5........eq.4, substitute b,c, and d

50.4993a+-+4428.45=49.153a+-+4473.58
50.4993a+-+49.153a=4428.45+-+4473.58
1.3463a=-45.13
a=-45.13%2F1.3463
a=+-33.52

go to
b+=+355.064+-+12.4999a...............3), substitute a
b+=+355.064+-+12.4999%2A-33.52
b=774.06

go to
d=15392.8+-+65.9988a........................5), substitute a
d=15392.8+-+65.9988%2A-33.52
d=17605.08

go to
c=5303%2F3-9%28-33.52%29-3%2A774.06-17605.08%2F3...........eq.1, substitute a, b,and d
c=+-6121.19

solutions:
a-33.52, b774.06, c-6121.19, d17605.1

the regression equation is:
y=-33.52x%5E3%2B774.06x%5E2-6121.19x%2B17605.08


b. Use the equation to find the number of boxes, to the nearest whole number, that the team will sell if they charge $4.35 per box.

y=-33.52x%5E3%2B774.06x%5E2-6121.19x%2B17605.08....plug in 4.35+for x
y=-33.52%2A4.35%5E3%2B774.06%2A4.35%5E2-6121.19%2A4.35%2B17605.08
y=2865.93

c. One year, the team only sold 125 boxes of chocolates. What price did they charge for each box?

y=-33.52x%5E3%2B774.06x%5E2-6121.19x%2B17605.08....plug in 125 for+y
125=-33.52x%5E3%2B774.06x%5E2-6121.19x%2B17605.08.....using calculator
->there is one real solution and it is x10.2

d. If the team raises the price too high, they will not sell any boxes. Use your regression equation to predict the price of a box that will result in zero boxes sold.
0=-33.52x%5E3%2B774.06x%5E2-6121.19x%2B17605.08.....using calculator
->there is one real solution and it is
x10.36