Question

QUESTION INVOLVING MATRICIES Scientific studies suggest that some animals regulate their intake of different types of food available in the environment to achieve a balance between the proportion, and ultimately the total amount, of macro-nutrients consumed. Macro-nutrients are categorised as protein, carbohydrate or fat/lipid. A seminal study on the macro-nutrient intake of migratory locust nymphs (Locusta migratoria) suggested that the locust nymphs studied sought and ate combinations of food that balanced the intake of protein to carbohydrate in a ratio of 45:55 [1]. Assume that a locust nymph finds itself in an environment where only two sources of food are available, identified as food X and food Y . Food X is 20% protein and 80% carbohydrate, whereas food Y is 70% protein and 30% carbohydrate. Assuming that the locust eats exactly 100 mg of food per day, determine how many milligrams of food X and food Y the locust needs to eat per day to reach the desired intake balance between protein and carbohydrate. [1] D Raubenheimer and S J Simpson, The geometry of compensatory feeding in the locust, Animal Behaviour

Answer 1

Let us assume that the Quantities of food X and Y eaten by the locusts be p & q mg respectively.

And, let the sum of the daily intake Carbs & Protien be r.

Therefore,

p+q=100 .....Eqn (1)

0.2p+0.8q=0.45r i.e 20p+80q=45r ......Eqn(2)

0.7p+0.3q=0.55r i.e 70p+30q=55r ......Eqn(3)

Now, to solve these Eqn, Multiply Eqn(2) with 55 and Eqn(3) with 45

Therfore,

1100p+4400q=2475r .....Eqn(4)

3150p+1350q=2475r .....Eqn(5)

Now, we substract Eqn(4) from Eqn(5) and get:

2050p-3050q=0 .....Eqn(6)

From Eqn(1), we have q=100-p

Substitung this value in Eqn(6):

2050p-3050(100-p)=0

i.e. 2050p-30500-3050p=0

i.e. 1000p=30500

p = 30.5

Therefore, q= 100- 30.5

q= 69.5

Thus, the locusts need to eat 30.5 mg of Food x and 69.5 mg of Food Y daily to reach the desired intake balance between protien and carbohydrates

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