Linear Systems of Equations

Question

Yummy Nuts has an inventory of 65,000 ounces of peanuts, the same amount of almonds, and 46,000 ounces of cashews, from which it will manufacture three nut mixes. Its regular mix contains 10 oz of peanuts, and 3 oz each of almonds and cashews. Its special mix contains 6 oz of peanuts, 6 oz of almonds and 4 oz of cashews, while its deluxe mix contains no peanuts, 10 oz of almonds and 6 oz of cashews. How many items of each type of mix should the company produce if it wants to deplete its entire inventory?

Answer 1

Hi Keira,

Here's how to tackle this type of prob.

Let R = number of regular mixes manufactured

Let S = number of special mixes manufactured

Let D = number of deluxe mixes manufactured

Then, ounces of peanuts used = 10R + 6S + 0D

And, ounces of almonds used = 3R + 6S + 10D

And, ounces of cashews used = 3R + 4S + 6D

=> 10R +6S <= 65000 . . . . . . . . . . . . . . (1)

and 3R +6S+ 10D <= 65000 . . . . . . . . . . . . . . (2)

and 3R + 4S + 6D <= 46000 . . . . . . . . . . . . . . (3)

Now, solve these 3 simultaneous equations in the usual way, to find R, S, and D.

Answer 2

Given information:

65,000 oz. peanuts

65,000 oz. almonds

46,000 oz. cashews

Mixture 1: 10 oz. P + 3 oz. A + 3 oz. C

Mixture 2: 6 oz. P + 6 oz. A + 4 oz. C

Mixture 3: 0 oz. P +10 oz. A + 6 oz C

When we make all of the mixtures we can, the total number of ingredients should add up to what our totals are.

Let a be the number of mixture 1 made, b be the number of mixture 2 made, and c be the number of mixture 3 made

a*M1+b*M2+c*M3 = 65k oz P + 65k oz A + 46k oz C

a(10P+3A+3C) + b(6P+6A+4C) + c(0P+10A+6C) = 65k P + 65k A + 46k C

10aP + 3aA + 3aC + 6bP + 6bA + 4bC +0cP + 10cA +6cC = 65k oz P + 65k oz A + 46k oz C

Rearrange this to get all P together, all A together and all C together.

(10a+6b+0c)P + (3a+6b+10c)A + (3a+4b+6c)C = 65k oz P + 65k oz A + 46k oz C

Now, we can say that the following equations must be true:

10a+6b+0c = 65000

3a+6b+10c = 65000

3a+4b+6c = 46000

Now, reduce an augmented matrix to REF (upper triangular matrix with 1s in the pivots)

eventually we get the following matrix:

[1 .6 0 | 6500 ]

[0 1 50/21 | 32500/3]

[0 0 1 | 3500 ]

By back-solving, we get that a = 5000, b = 2500, c = 3500.

So, we need 5000 of mixture 1, 2500 of mixture 2, and 3500 of mixture 3!