Question
Question: The maximum value of \(µ = 3x + 4y\) subjected to the conditions \(x + y \leq 40,x + 2y \leq 60;x,y ...
The maximum value of µ=3x+4y subjected to the conditions x+y≤40,x+2y≤60;x,y≥0 is
A
130
B
120
C
40
D
140
Answer
140
Explanation
Solution
Obviously Max µ=3x+4yat(20,20)µ=60+80=140.
