Find the maximum value of 4x+y if:
3x+y ≤ 31
2x+y ≤ 23
2x-y ≤ 14
x+6y ≥ 20
y ≤ 11
y ≥ 0
x ≥ 2
The max will occur at a "corner point" but this is very hard to identify with a graph because there are so many "overlapping" graphs....
WolframAlpha gives the following solution :
https://www.wolframalpha.com/input/?i=maximize+4x+%2By+given+3x%2By++%E2%89%A4+31+,++2x%2By++%E2%89%A4+23,+++2x-y++%E2%89%A4+14,+++x%2B6y++%E2%89%A5+20,+++y+%E2%89%A4+11,+++y+%E2%89%A5+0,+++x+%E2%89%A5+2
It appears that the max occurs at (9, 4) and the max is 4(9) + 4 = 40
Here's the Desmos graph : https://www.desmos.com/calculator/opl1cuchfu
Maybe you can find the max corner point much easier than I can.......!!!!!
The max will occur at a "corner point" but this is very hard to identify with a graph because there are so many "overlapping" graphs....
WolframAlpha gives the following solution :
https://www.wolframalpha.com/input/?i=maximize+4x+%2By+given+3x%2By++%E2%89%A4+31+,++2x%2By++%E2%89%A4+23,+++2x-y++%E2%89%A4+14,+++x%2B6y++%E2%89%A5+20,+++y+%E2%89%A4+11,+++y+%E2%89%A5+0,+++x+%E2%89%A5+2
It appears that the max occurs at (9, 4) and the max is 4(9) + 4 = 40
Here's the Desmos graph : https://www.desmos.com/calculator/opl1cuchfu
Maybe you can find the max corner point much easier than I can.......!!!!!
Let's look at this again......and we will remove the last 3 constraints.....these three just tell us that the solution will occur in the first quadrant where x ≥ 2 and y ≤ 11
Look at the graph now : https://www.desmos.com/calculator/upxf7i7tbd
If I read the graph correctly.....[a real leap of faith]...there appear to be only two corner points that satisfy the constraints....they are
(8,2) and (9,4).......but only one will maximize 4x + y....and that occurs at (9,4)