BITSAT 2014
- Exam
- BITSAT
- Year
- 2014
- Subject
- Mathematics
- Chapter
- Linear Programming
BITSAT 2014 · Mathematics · Linear Programming
Consider (x)/(2)+(y)/(4) ≥ 1 and (x)/(3)+(y)/(2) ≤ 1, x, y ≥ 0 ....
Consider and . Then number of possible solutions are :

Options
- AZero
- BUnique
- CInfinite
- DNone of these
Correct answer
C. Infinite
Step-by-step solution
Consider , x, convert them into equation and solve them and draw the graph of these equations we get and 
From graph region is finite but numbers of possible solutions are infinite because for different values of and we have different or infinite no. of solutions.

From graph region is finite but numbers of possible solutions are infinite because for different values of and we have different or infinite no. of solutions.