Vitamin, protein and fat are nutrients in spinach and groundnuts. Spinach contains 20, 40 and 30 units of vitamin, protein and fat respectively. Groundnuts contain 50, 20 and 40 un... Vitamin, protein and fat are nutrients in spinach and groundnuts. Spinach contains 20, 40 and 30 units of vitamin, protein and fat respectively. Groundnuts contain 50, 20 and 40 units of vitamin, protein and fat respectively. The cost of spinach per kilogram is shs. 5000. The cost of groundnuts per kilogram is shs. 3000. If the minimum requirement is 800 units of vitamin, 1000 units of protein and 1200 units of fat, how much spinach and groundnuts should be bought to satisfy the conditions? Read more

Steps to Solve

1. Define variables

Let $x$ be the amount of spinach in kilograms and $y$ be the amount of groundnuts in kilograms.

2. Formulate the objective function

The objective is to minimize the cost, which is given by: $$C = 5000x + 3000y$$

3. Formulate the constraints

Based on the minimum requirements for vitamin, protein, and fat, we have the following constraints: Vitamin: $20x + 50y \ge 800$ Protein: $40x + 20y \ge 1000$ Fat: $30x + 40y \ge 1200$ Also, $x \ge 0$ and $y \ge 0$ since we cannot have negative amounts of spinach or groundnuts.

4. Simplify the constraints

Simplify the inequalities: Vitamin: $2x + 5y \ge 80$ Protein: $2x + y \ge 50$ Fat: $3x + 4y \ge 120$

5. Solve the system of inequalities graphically or using linear programming techniques

First, find the intersection of $2x + 5y = 80$ and $2x + y = 50$: Subtract the second equation from the first to get $4y = 30$, so $y = 7.5$. Then, $2x + 7.5 = 50$, so $2x = 42.5$, and $x = 21.25$. Intersection point: $(21.25, 7.5)$

Next, find the intersection of $2x + 5y = 80$ and $3x + 4y = 120$: Multiply the first equation by 3 and the second by 2: $6x + 15y = 240$ $6x + 8y = 240$ Subtract the second equation from the first to get $7y = 0$, so $y = 0$. Then, $2x = 80$, so $x = 40$. Intersection point: $(40, 0)$

Next, find the intersection of $2x + y = 50$ and $3x + 4y = 120$: Multiply the first equation by 4 to get $8x + 4y = 200$ Subtract the third equation from this to get $5x = 80$, so $x = 16$. Then, $2(16) + y = 50$, so $32 + y = 50$, and $y = 18$. Intersection point: $(16, 18)$

6. Evaluate the objective function at the corner points of the feasible region

We need to consider the corner points of the feasible region. The corner points are $(16,18)$, $(40, 0)$, and $(21.25, 7.5)$.

Evaluate $C = 5000x + 3000y$ at these points: At $(16, 18)$: $C = 5000(16) + 3000(18) = 80000 + 54000 = 134000$ At $(40, 0)$: $C = 5000(40) + 3000(0) = 200000$ At $(21.25, 7.5)$: $C = 5000(21.25) + 3000(7.5) = 106250 + 22500 = 128750$

7. Consider other intersection points

We must also check $x=0$ and $y=0$ intercepts implied by the constraints to determine which points give the absolute minimum.

If $x=0$: $5y \ge 80 \implies y \ge 16$ $y \ge 50$
$4y \ge 120 \implies y \ge 30$

Therefore the $x=0$ intercept with the feasible region is $(0,50)$ giving a cost of $C=50*3000 = 150000$.

If $y=0$: $2x \ge 80 \implies x \ge 40$ $2x \ge 50 \implies x \ge 25$ $3x \ge 120 \implies x \ge 40$

Therefore the $y=0$ intercept with the feasible region is $(40,0)$ with a cost of $C = 40*5000 = 200000$.

8. Determine the minimum cost

The minimum cost is 128750 at $(21.25, 7.5)$. Since we are likely dealing with physical quantities, we might want to consider integer solutions close to this point. However, without further constraints or information, we will take the non-integer solution as the optimum.