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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+4yµ = 3x + 4y subjected to the conditions x+y40,x+2y60;x,y0x + y \leq 40,x + 2y \leq 60;x,y \geq 0 is

A

130

B

120

C

40

D

140

Answer

140

Explanation

Solution

Obviously Max µ=3x+4yat(20,20)µ=60+80=140µ = 3x + 4y\text{at}(20,20)µ = 60 + 80 = 140.