CBSE Class 12 · 2007 · Linear ProgrammingLong Answer

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If a young man rides his motorcycle at 25 km/hour, he has to spend Rs. 2 per km on petrol. If he rides at a faster speed of 40 km/hour, the petrol cost increases to Rs. 5 per km. He has Rs. 100 to spend on petrol and wishes to find what is the maximum distance he can travel within one hour. Express this as an LPP and solve it graphically.

Answer: Max distance = 30 km (50/3 km at 25 km/h and 40/3 km at 40 km/h)

Solution

Let

x

km be covered at 25 km/h and

y

km at 40 km/h.
Maximise distance:

Z = x + y

.
Petrol cost:

2 x + 5 y \leq 100

Time (at most 1 hour):

\frac{x}{25} + \frac{y}{40} \leq 1 ⟹ 8 x + 5 y \leq 200

x \geq 0, y \geq 0

Corner points:

(0, 0)

(25, 0)

(\frac{50}{3}, \frac{40}{3})

(intersection of

2 x + 5 y = 100

and

8 x + 5 y = 200

) and

(0, 20)

.
Intersection:

8 x + 5 y = 200

minus

2 x + 5 y = 100

gives

6 x = 100 ⟹ x = \frac{50}{3}

, then

y = \frac{40}{3}

.
Evaluate

Z = x + y

(25, 0)

Z = 25

(\frac{50}{3}, \frac{40}{3})

Z = \frac{90}{3} = 30

← maximum

(0, 20)

Z = 20

∴ Z_{m a x} = 30 km at (\frac{50}{3}, \frac{40}{3})

Maximum distance is 30 km:

\frac{50}{3}

km at 25 km/h and

\frac{40}{3}

km at 40 km/h.

CBSE Class 12 Board · 2007 · Subjective

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CBSE Class 12 · 2007 · Linear ProgrammingLong Answer

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