Question 1148051
x = the number of kilograms of beef.
y = the number of kilograms of pork.
z = the number of kilograms of fat.


your constraint equations are:
x + y + z = 50
z <= 10
y <= .25 * x
x >= 0
y >= 0


your objective function is:
cost = 6 * x + 4 * y + 2 * z


if you are dealing with a two dimensional graph, then you have to find a way to reduce the number of variables to two instead of 3.


you can use some logic here.
since the cost of fat is the cheapest, it makes sense that you would want to maximize the use of fat.
that means that you would use 10 kilograms of fat in whatever mixture you need.


therefore, your objective function becomes cost = 
6 * x + 4 * y + 20


a little logic in your constraint equations will allow you to eliminate z from those as well.


if z <= 10, that  means the x + y >= 40.
you can replace z <= 10 with x + y >= 40.


your constraint equations become:
x + y = 40
x + y >= 40
y <= .25 * x
x >= 0
y >= 0


those top two constraint equations are conflicting, but you will get the same answer whether you use one or the other.
i'll do it all three ways to show you the answer becomes the same regardless of which way you choose to do it.


using the desmos.com calculator, you would graph the opposite of these constraints.
the area of the graph that is not shaded is your region of feasibility.
the corner points of that region are where your least cost solution lies.


first graph uses all the constraint inequality opposites.
you would graph.
x + y = 40]
x + y <= 40
x >= .25 * x
x <= 0
y <= 0


that graph looks like this:


<img src = "http://theo.x10hosting.com/2019/110301.jpg" alt="$$$" >


second graph uses x + y <= 40 and not x + y = 40:
x + y <= 40
x >= .25 * x
x <= 0
y <= 0


<img src = "http://theo.x10hosting.com/2019/110302.jpg" alt="$$$" >


third graph uses x + y = 40 and not x + y <= 40:
x + y = 40
x >= .25 * x
x <= 0
y <= 0


<img src = "http://theo.x10hosting.com/2019/110303.jpg" alt="$$$" >


in the third graph, the solution has to be on the line x + y = 40.


don't forget that we are graphing the opposite of the inequalities and the region of feasibility becomes the area of the graph that is not shaded.


you will see that all graphs point to two possible solutions.
they are:
(x,y) = (32,8) or (x,y) = (40,0)


your objective function is cost = 6 * x + 4 * y + 20


at your two possible low cost points, the objective function is evaluated to get:


when x = 32 and y = 8, the cost is equal to 32 * 6 + 8 * 4 + 20 = 244


when x = 40 and y = 0, the cost is equal to 40 * 6 + 0 * 4 + 20 = 260.


you have a lowest cost solution when x = 32 and y = 8.
the lowest cost is equal to 244.


all constraint requirements are met, as they have to be.
x >= 0 becomes 32 >= 0 which is true.
y >= 0 becomes 8 >= 0 which is true.
y <= .25 * x becomes 8 <= .25 * 32 which becomes 8 <= 8 which is true.


i used a linear programming simplex tool to confirm this would be the least possible cost solution if we used all three variables.


the results from the linear programming simplex tool confirmed the solution is correct.


a picture of the results from that tool is shown below.


<img src = "http://theo.x10hosting.com/2019/110304.jpg" alt="$$$" >


the linear equation simplex tool couldn't handle y <= .25x, so i had to change that inequality to y - .25x <= 0 by subtracting .25x from both sides of it.


i included x >= 0 and y >= 0, but the simplex tool didn't require that, so i could have left them out and gotten the same result.


the simplex tool confiirmed that the low cost solution is when x = 32 and y = 8.


that's your solution.


your lowest cost will be when you use 32 kilograms of beef and 8 kilograms of pork and 10 kilogramas of fat the mixture.