6: Real World Constraints

Solution

Part 1

Let x, y and z be the number of pieces of chocolate cake, hard boiled eggs, and serves of Nutri-grain, respectively.

We have these equations:

2.5x + 6y + 9z = 50

35x + 0.5y + 85z = 310

10x + 5y + z = 70

Solving this system (using a calculator), (x,y,z) = (4.89, 3.88, 1.61). However, Jericho could probably not eat these exact numbers.

If he ate 5 pieces of chocolate cake, 4 hard-boiled eggs, and 1.6 serves of Nutri-grain, he would get very close.

Part 2

The equations change to:

2.5x + 6y + 9z = 150

35x + 0.5y + 85z = 200

10x + 5y + z = 50

The planes represented by these equations have moved to be new parallel planes. So we expect there to still be a unique solution.

Solving this system (using a calculator), (x,y,z) = (-5.7, 20.5, 4.6).

Although there is a unique solution, it is not possible for Elspeth to use these foods in her diet because she cannot eat -5.7 pieces of chocolate cake. (We expect x, y and z to be non-negative).

Part 3

Let x, y and z be the number of snack bars, tasty meals and healthy meals Robbie eats. The equations are now:

(protien): 6x + 14y + 10z = 50

(carbs): 20x + 80y + 100z = 300

(fat): 10x + 20y + 10z = 70

Using elimination, we find that this system of equations simplifies to:

0.3(carbs) - (protien): 10y + 20z = 40

(carbs) - 2(fat): 40y + 80z = 160

from which we can see the system is consistent.

If we let z be a real number t, then

z = t

y = 4 - 2z = 4 - 2t

x = 7 - 2y - z = 3t - 1

So (x,y,z) = (3t -1, 4 - 2t, t).

Since these must all be positive numbers, we need 3t - 1 ≥ 0, 4 - 2t ≥ 0 and t ≥ 0.

So t ≥ 0.333, t ≤ 2 and t ≥ 0.

The range is 0.333 ≤ t ≤ 2.

This could be (0, 3.33, 0.33), which might not suit Robbie as it is not a whole number of pre-packaged meals.

He could eat (2,2,1) or (5,0,2) from these options with whole-numbers of the products.

We suggest, for variety, Robbie eats two Snack Bars, two Tasty Meals, and one Healthy Meal.