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Q.

To maintain his health, a person must fulfill certain minimum daily requirements for several kinds of nutrients. Assuming that there are only three kinds of nutrients — calcium, protein and Calories — and the person's diet consist of only two food items 1 and 2, whose price and nutrient contents are shown in the table below:

NutrientsFood I (per lb)Food II (per lb)Minimum daily requirement
Calcium10420
Protein5520
Calories2613
Price (in Rs.)0.601.00

What combination of two food items will satisfy the daily requirement and entail the least cost? Formulate this problem as a LPP.

Sikkim CbseNCERTSubjective· 3mImportance★★★★★
64% · 14/22 Questions
✓ Free question

LPP: minimise Z=0.60x+1.00yZ = 0.60x + 1.00y subject to the three nutrient constraints. The least-cost diet is 2.75 lb of Food I and 1.25 lb of Food II, costing Rs. 2.90.

Formulation

Let xx = pounds of Food I and yy = pounds of Food II, with x≥0, y≥0x\ge0,\ y\ge0.

Minimise (cost):

Z=0.60x+1.00yZ = 0.60x + 1.00y

Subject to (each nutrient must meet its minimum daily requirement):

10x+4y≥20(calcium)5x+5y≥20(protein)2x+6y≥13(calories)x, y≥0\begin{aligned} 10x + 4y &\ge 20 &&\text{(calcium)}\\ 5x + 5y &\ge 20 &&\text{(protein)}\\ 2x + 6y &\ge 13 &&\text{(calories)}\\ x,\,y &\ge 0 \end{aligned}

Solving graphically

The feasible region is unbounded (above all three lines). Its lower-boundary corner points and their costs:

Corner pointHow obtainedCost Z=0.60x+1.00yZ = 0.60x + 1.00y
(0, 5)(0,\,5)calcium ∩ yy-axis5.005.00
(23, 103)\left(\tfrac23,\,\tfrac{10}{3}\right)calcium ∩ protein3.733.73
(2.75, 1.25)(2.75,\,1.25)protein ∩ calories2.90\mathbf{2.90}
(6.5, 0)(6.5,\,0)calories ∩ xx-axis3.903.90

(The calcium ∩ calories point (1.31, 1.73)(1.31,\,1.73) fails protein, since x+y=3.04<4x+y = 3.04 < 4, so it is not a feasible vertex.)

The point (2.75, 1.25)(2.75,\,1.25) satisfies all constraints — calcium 32.5≥2032.5\ge20, protein 20≥2020\ge20, calories 13≥1313\ge13 — at the lowest cost:

Z=0.60(2.75)+1.00(1.25)=1.65+1.25=Rs. 2.90.Z = 0.60(2.75) + 1.00(1.25) = 1.65 + 1.25 = \text{Rs. } 2.90.

✓Final answer

Minimise Z=0.60x+1.00yZ = 0.60x + 1.00y subject to 10x+4y≥2010x+4y\ge20, 5x+5y≥205x+5y\ge20, 2x+6y≥132x+6y\ge13, x,y≥0x,y\ge0. The least-cost combination is 2.75 lb of Food I and 1.25 lb of Food II, at a minimum cost of Rs. 2.90.

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