Jimmy’s little brother Timmy likes to collect stuffed animals. Stuffed animals are stuffed with 2 different things, cotton and beads. He always wants to have at least 10 stuffed animals. He has a special shelf to place his stuffed animals. The shelf can hold 30 lbs. at most and has 504 square inches of shelf space. Timmy has $120 dollars to buy stuffed animals with. Each cotton stuffed animal is $6 and each bead stuffed animal is $4. Each cotton stuffed animal weighs half a pound and takes up 21 square inches of space. Each bead stuffed animal weighs 2 pounds and takes up 24 square inches of space.
What is the maximum number of stuffed animals Timmy can buy?
To first start this problem we had to define our variables. The variables used in this problem were B, for the number of stuffed animals stuffed with beads, and C, for the number of stuffed animals stuffed with cotton.
Secondly we needed to identify our constraints. Since we know that it is not possible for Timmy to have less than 0 stuffed animals, the first two constraints were B>_0 and C>_0. When these inequalities are shown C is on the x-axis and B is on the y-axis, and the whole first quadrant is shaded. I labeled both of these the minimum on the graph.
Another constraint given by the problem above was that he wants at least 10 stuffed animals at all times. Which means that the stuffed animals can either be filled with cotton or beads. The inequality for this was C+B>_10.
Timmy’s shelf can only hold 30 pounds. Since the cotton stuffed animals are only ½ pounds and the bead stuffed animals are 2 pounds, the constraint can be given as ½ C+2B>_30. I labeled this weight on the graph.
The problem also tells us that Timmy’s shelf has a limited amount of space of 504 square inches. The Beaded stuffed animals take up 24 squares inches while the cotton stuffed animals take up 21 square inches. The inequality for this was 21C+24B