CBSE Class 12 · 2009 · Linear ProgrammingLong Answer

A diet is to contain at least 80 units of Vitamin A and 100 units of minerals. Two foods

F_{1}

and

F_{2}

are available. Food

F_{1}

costs Rs. 4 per unit and

F_{2}

costs Rs. 6 per unit. One unit of food

F_{1}

contains 3 units of Vitamin A and 4 units of minerals. One unit of food

F_{2}

contains 6 units of Vitamin A and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for a diet that consists of a mixture of these two foods and also meets the minimal nutritional requirements.

Answer: Min cost = Rs 104 at (24, 4/3): 24 units of F1 and 4/3 units of F2

Solution

Let the diet contain

x

units of

F_{1}

and

y

units of

F_{2}

.
Cost to minimise:

Z = 4 x + 6 y

.
Vitamin A:

3 x + 6 y \geq 80

Minerals:

4 x + 3 y \geq 100

x \geq 0, y \geq 0

Corner points of the unbounded feasible region:

A (0, \frac{100}{3})

B (24, \frac{4}{3})

(intersection of

3 x + 6 y = 80

and

4 x + 3 y = 100

) and

C (\frac{80}{3}, 0)

.
Evaluate

Z = 4 x + 6 y

A (0, \frac{100}{3})

Z = 200

B (24, \frac{4}{3})

Z = 96 + 8 = 104

← minimum

C (\frac{80}{3}, 0)

Z = \frac{320}{3} \approx 106.67

∴ Z_{m i n} = Rs 104 at (24, \frac{4}{3})

The minimum cost is Rs 104, using 24 units of

F_{1}

and

\frac{4}{3}

units of

F_{2}

Structure Unbounded feasible region for the diet problem, with corner points (0, 100/3), (24, 4/3) and (80/3, 0). — Unbounded feasible region for the diet problem, with corner points (0, 100/3), (24, 4/3) and (80/3, 0).

CBSE Class 12 Board · 2009 · Subjective

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CBSE Class 12 · 2009 · Linear ProgrammingLong Answer

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