Question

A dietician has to develop a special diets using two foods X and Y. Each packet (containing 30g) of food. X contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Y contains 3units of calcium , 20 unit of iron, 4 units of cholesterol and 3 units of vitamin A.The diet requires atleast 240 units of calcium, atleast 460 units of iron and atmost 300 units of cholesterol. The corner points of the feasible region are

• A. (2,72), (40,15), (15,20)
• B. (0,23), (40,15), (2,72)
• C. (2,72), (15,20), (0,23)
• D. (2,72), (40,15), (115,0)

Answer 1

The Correct Option is A

Let the quantity of food X used as x packets and the quantity of food Y used as y packets.
The constraints can be represented as follows: 
\(12x + 3y ≥ 240\) (constraint for calcium) \(4x + 20y ≥ 460\) (constraint for iron) \(6x + 4y ≤ 300\) (constraint for cholesterol) 
To find the corner points, we solve these constraints as a system of linear inequalities. 
First, let's plot the feasible region on a graph: 

The constraints can be rearranged to isolate y: 
\(y \geq \frac{240 - 12x}{3}\) (constraint for calcium) 
\(y \geq \frac{460 - 4x}{20}\) (constraint for iron) 
\(y \leq \frac{300 - 6x}{4}\)(constraint for cholesterol) 
By plotting these inequalities on a graph, we can find the corner points where the feasible region intersects. 
After graphing, we find that the corner points of the feasible region are approximately (2, 72), (40, 15), and (15, 20). 
Therefore, the correct option is (A) (2, 72), (40, 15), (15, 20).

Answer 2

Let the number of packets of food X = x
Let the number of packets of food Y = y

The nutrient constraints are formed as follows:

• Calcium: \(12x + 3y \geq 240\)
• Iron: \(4x + 20y \geq 460\)
• Cholesterol: \(6x + 4y \leq 300\)
• Non-negativity: \(x \geq 0, y \geq 0\)

Solving this system graphically or using LPP methods, we find the feasible region bounded by the intersection points of the constraint lines.

By solving pairs of equations, we get the corner points of the feasible region as: (2,72), (40,15), (15,20)

 

Correct Answer: (2,72), (40,15), (15,20)

Answer 3

Let \(x\) be the number of packets of food X and \(y\) be the number of packets of food Y.

We need to formulate the constraints based on the nutritional requirements and limits.

Constraints: 

1. Calcium: At least 240 units. Food X has 12 units/packet, Food Y has 3 units/packet. \[ 12x + 3y \ge 240 \] Dividing by 3, we get: \[ \mathbf{4x + y \ge 80} \quad \dots (1) \]
2. Iron: At least 460 units. Food X has 4 units/packet, Food Y has 20 units/packet. \[ 4x + 20y \ge 460 \] Dividing by 4, we get: \[ \mathbf{x + 5y \ge 115} \quad \dots (2) \]
3. Cholesterol: At most 300 units. Food X has 6 units/packet, Food Y has 4 units/packet. \[ 6x + 4y \le 300 \] Dividing by 2, we get: \[ \mathbf{3x + 2y \le 150} \quad \dots (3) \] (Note: Vitamin A information is provided but no requirement is stated, so it doesn't form a constraint.)
4. Non-negativity: The number of packets cannot be negative. \[ \mathbf{x \ge 0, y \ge 0} \quad \dots (4) \]

The corner points of the feasible region are the intersection points of the boundary lines of these inequalities, provided they satisfy all constraints.

Boundary lines:

• L1: \(4x + y = 80\)
• L2: \(x + 5y = 115\)
• L3: \(3x + 2y = 150\)

We find the intersection points:

• Intersection of L1 and L2: From L1, \(y = 80 - 4x\). Substitute into L2: \(x + 5(80 - 4x) = 115\) \(x + 400 - 20x = 115\) \(-19x = -285 \implies x = 15\) \(y = 80 - 4(15) = 80 - 60 = 20\) Point: (15, 20). Check feasibility: (1) \(4(15)+20=80 \ge 80\), (2) \(15+5(20)=115 \ge 115\), (3) \(3(15)+2(20)=45+40=85 \le 150\). Point is feasible.
• Intersection of L1 and L3: From L1, \(y = 80 - 4x\). Substitute into L3: \(3x + 2(80 - 4x) = 150\) \(3x + 160 - 8x = 150\) \(-5x = -10 \implies x = 2\) \(y = 80 - 4(2) = 80 - 8 = 72\) Point: (2, 72). Check feasibility: (1) \(4(2)+72=80 \ge 80\), (2) \(2+5(72)=2+360=362 \ge 115\), (3) \(3(2)+2(72)=6+144=150 \le 150\). Point is feasible.
• Intersection of L2 and L3: From L2, \(x = 115 - 5y\). Substitute into L3: \(3(115 - 5y) + 2y = 150\) \(345 - 15y + 2y = 150\) \(-13y = -195 \implies y = 15\) \(x = 115 - 5(15) = 115 - 75 = 40\) Point: (40, 15). Check feasibility: (1) \(4(40)+15=175 \ge 80\), (2) \(40+5(15)=115 \ge 115\), (3) \(3(40)+2(15)=120+30=150 \le 150\). Point is feasible.

We also need to consider intersections with the axes (x=0 and y=0), but as shown in the thought process, these intersections do not yield feasible points satisfying all constraints simultaneously.

The feasible corner points are (15, 20), (2, 72), and (40, 15).

Comparing this set of points with the given options:

• (2,72), (40,15), (15,20) - Matches our findings.
• (0,23), (40,15), (2,72) - Includes (0,23) which is not feasible (violates \(4x+y \ge 80\)).
• (2,72), (15,20), (0,23) - Includes (0,23) which is not feasible.
• (2,72), (40,15), (115,0) - Includes (115,0) which is not feasible (violates \(3x+2y \le 150\)).

Therefore, the correct set of corner points for the feasible region is:

(2,72), (40,15), (15,20)