The Candy Problem

Presented by:

Dana TeCroney

Daniel bought one pound of jellybeans and two pounds of chocolates for $2.00. A week later, he bought four pounds of caramels and one pound of jellybeans, paying $3.00. The next week, he bought three pounds of licorice, one pound of jellybeans and one pound of caramels for $1.50. How much would he have to pay on his next trip to the candy store, if he bought one pound of each of the four candies?

In the following model, let

L = amount of licorice (lbs)

C = amount of chocolate (lbs)

K = amount of caramels (lbs)

J - amount of Jelly Beans (lbs)

X = the cost of one pound of each candy

Using these variables, the following equations can be set up:

Method I: Matrices

Matrices can be used to show that this system does not have a unique solution. If you can remember linear algebra, one common theorem was that if a matrix can be reduced to have a row of all zeros, then the solution of not unique.

Here is the coefficient matrix:

Now, we reduce this by performing the operation row(4) - 3row(1):

Two more operations get us were we need to be: row(4) + 3/2 row(2) and row(4) + ½row(3).

As you can see, the row(4) is all zeros, hence this system will not have a unique solution. So is all lost?

Not really, the last row implies that 6 + 3X = 0, so X = 2. If you look back to the system, X was the cost of one pound of each candy, just want we were looking for.

Method II: Equations

Recall, our system was:

Consider this system if the second equation was multiplied 3 and the fourth equation by 2, then the following results (just considering the bottom three equations)

After multiplying the equations, they can be added and will have a common multiple of 2. Divide through by 2 and you can see that the sum of one pound of each of the candies is $2.00.

Possible Extensions:

Can you figure out what some possible costs for each pound of candy would be?