SOLUTION: Rebecca wants to bake cookies and cupcakes for a bake sale. She can bake 15 cookies at a time and 12 cupcakes at a time. She needs to make at least 20 baked goods but no more than

Algebra ->  Graphs -> SOLUTION: Rebecca wants to bake cookies and cupcakes for a bake sale. She can bake 15 cookies at a time and 12 cupcakes at a time. She needs to make at least 20 baked goods but no more than       Log On


   



Question 231252: Rebecca wants to bake cookies and cupcakes for a bake sale. She can bake 15 cookies at a time and 12 cupcakes at a time. She needs to make at least 20 baked goods but no more than 360, and she wants to have at least three times as many cookies as cupcakes. What combination of batches of each could Rebecca make?
I need to get two system of inequality equations from this but I am not sure if I am getting the right equations. The equations I have are: x >= 3y and 120 <= x+y <=360

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = number of batches of cookies she could make
Let b = number of batches of cupcakes she could make
given:
20+%3C=+15a+%2B+12b+%3C=+360
15a+%3E=+3%2A12b
15a+%3E=+36b
-----------------
15a+%2B+12b+%3E=+20
15a+%3E=+20+-+12b
(1) a+%3E=+-%284%2F5%29%2Ab+%2B+4%2F3
and
15a+%2B+12b+%3C=+360
15a+%3C=+360+-+12b
(2) a+%3C=+-%284%2F5%29%2Ab+%2B+24
and
(3) a+%3E=+%2812%2F5%29%2Ab
If I plot a on vertical, b on horizontal for these 3 equations, I get

It looks like the solution is the small triangular
region bounded by the 3 lines and the a-axis
If I try a=18 and b=6, then
20+%3C=+15a+%2B+12b+%3C=+360
20+%3C=+15%2A18+%2B+12%2A6+%3C=+360
20+%3C=+270+%2B+72+%3C=+360
20+%3C=+342+%3C=+360
OK, and
15a+%3E=+36b
5a+%3E=+12b
5%2A18+%3E=+12%2A6
90+%3E=+72
Hope I got it right- I think so