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Refer to question 15. Determine the maximum distance that the man can travel.

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प्रश्न

Refer to question 15. Determine the maximum distance that the man can travel.

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उत्तर

Referring to the solution of Question No.15, we have

Maximise Z = x + y subject to the constraints

Let 2x + 3y = 120

x 0 60
y 40 0

Let 8x + 5y = 400

x 0 50
y 80 0


2x + 3y ≤ 120  ......(i)

8x + 5y ≤ 400  ......(ii)

x ≥ 0, y ≥ 0

On solving eq. (i) and (ii) we get

x = `300/7` and y = `80/7`

Here, OABC is the feasible region whose corner points are O(0, 0), A(50, 0), `"B"(300/7, 80/7)` and C(0, 40).

Let us evaluate the value of Z

Corner points Value of Z = x + y  
O(0, 0) Z = 0 + 0 = 0  
A(50, 0) Z = 50 + 0 = 50 km  
`"B"(300/7, 80/7)`

Z = `300/7 + 80/7`

= `380/7` = 54.3 km

← Maximum
C(0, 40) Z = 0 + 40 = 40 km  

Hence, the maximum distance that the man can travel is `54 2/7` km at `(300/7, 80/7)`.

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अध्याय 12: Linear Programming - Exercise [पृष्ठ २५३]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 12 Linear Programming
Exercise | Q 20 | पृष्ठ २५३

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