Appliction Problems on Systems of Equation

Question

A baker wishes to produce three types of cookies. In addition to the other ingredients, she mixes flour, milk and nuts with the following proportions:

A: 7 c flour, 5 c milk, 1 oz nuts

B: 3 c flour, 2 c milk, 2 oz nuts

C: 4 c flour, 3 c milk, 3 oz nuts

If 67 cups of flour, 48 cups of milk and 32 oz of nuts are available, how much of each kind of cookies can be made?

Answer 1

In order to solve this problem we need three equations to be true at the same time. Let A be the amount of stuff in recipe A, B the amount of stuff in recipe B, and C the amount of stuff in recipe C.

Given the values in the problem, we have 3 equations:

7A + 3B + 4C = 67 (flour)

5A + 2B + 3C = 48 (milk)

A + 2B + 3C = 32 (nuts)

Since we have 3 equations and 3 unknowns, we can solve this using any number of methods. I'm going to solve it by elimination and substitution.

If we take the flour equation, subtract the milk equation, and twice the nut equation, we can eliminate A entirely, since 7 - 5 - 2(1) = 0.

Algebraically what we do is:

7A + 3B + 4C = 67

-(5A + 2B + 3C = 48)

-2(A + 2B + 3C = 32)

Simplifying a bit, what we get is:

7A + 3B + 4C = 67

-5A - 2B - 3C = -48

-2A - 4B - 6C = -64

Further simplifying what we get is:

(7A - 5A - 2A) + (3B - 2B - 4B) + (4C - 3C - 6C) = 67 - 48 - 64

-3B + -5C = -45

3B + 5C = 45

We can solve this equation for either B or C. Let's solve for B.

3B = 45 - 5C

B = 15 - 5/3 C

We can also take the second two equations and actually solve for A directly:

5A + 2B + 3C = 48

-(A + 2B + 3C = 32)

Simplifying, what we get is:

(5A - A) + (2B - 2B)+ (3C - 3C) = 48 - 32

4A = 16, which implies A = 4.

Since we've solved for A and for B (in terms of C), now we can solve for C. Let's use our first equation, which was:

7A + 3B + 4C = 67

Substituting A = 4 and 3B = 45 - 5C, we get:

7(4) + 45 - 5C + 4C = 67

28 + 45 - C = 67

73 - C = 67, so C = 6.

Since we have B in terms of C, now we can solve for B:

B = 15 - 5/3 C = 15 - 5/3 (6) = 15 - 10 = 5

Therefore B = 5.

So now we have our solution: We can make 4 batches of recipe A, 5 batches of recipe B, and 6 batches of recipe C.