Question

The staff dietician at the California Institute of Trigonometry has to make up a meal with 600 calories, 20 grams of protein, and 200 milligrams of vitamin C. There are three food types to choose from: rubbery jello, dried fish sticks, and mystery meat. They have the following nutritional content per ounce. Jello Fish Sticks Mystery Meat Calories 10 50 200 Protein 1 3 .2 Vitamin C 30 10 o (a) Make a mathematical model of the dietician\'s problem with a system of three linear equations. (b) Find an approximate solution (accurate to within 10%)

Answer 1

Answer:

The amount of rubbery jello=5.038 ounce

Dried fish stick = 4.886 ounce

Mystery meat = 1.526 ounce.

Step-by-step explanation:

Solutions/Explanation: (a) Suppose the amount of rubbery jello = $x$ ounce,  

The dried fish sticks amount to be = $y$ ounce  

The amount of mystery meat = ounce.

Specification of the ingredients

Rubbery jello has: Calories=10 Cal/ounce, Protien=1 gram/ounce, Vitamin C= 30 mg/ounce

Dried fish sticks: Calories=50 Cal/ounce, Protien=3 grams/ounce, Vitamin C=10 mg/ounce

Mystery meat: Calories=200 Cal/ounce, Protien=0.2 gram/ounce, Vitamin C=0 mg/ounce,

Now,  

Total calories = $600$.  

Thus,

$10x+50y+200z=600\quad\Idot (1)$

The, total Protien = $20 g$.  

Thus,

$x+3y+0.2z=20 \quad\Idot (2)$

also, total Vitamin C = .

Thus,

$30x+10y+0z=200 \quad\Idot (3)$

Therefore,

1, 2 and 3 are the mathematical models of the dietician's problem with a system of three linear equations.

(b) Now, we know three equations:

$10x+50y+200z=600$

OR  

$x+5y+20=60 \quad\Idot (4)$

From eqn 2,  

$x+3y+0.2z=20 \quad\Idot (5)$

From eqn 3.  

$30x+10y=200$

OR  

$3x+y=20 \quad\Idot (6)$

On solving the eqn 4, 5 and 6 for $x, y$ and $z,$

Multiplying eqn 5 by 100,

$100x+300y+20z=2000 \quad\Idot (7)$

Now, subtract eqn 4 from eqn 7

$99x+295y=1940 \quad\Idot (8)$

Now, multiplying eqn 6 by 295. we have,

$885x+295y=5900 \quad\Idot (9)$

Now, subtract, eqn 8 from eqn 9, we get,

$786x=3960$

$x=\frac{3960}{786}$

 $x=5.038 \quad\Idot (10)$

Now, substituting into eqn 6, we get

$y=4.886 \quad\Idot (11)$

Now, substitute eqn 10 and 11 into eqn 4, we have

$z=1.526$