Question

Solve the following graphically,
Minimize z = 30x + 20y subject to:
 x + y ≤ 8,
 x + 2y ≥ 4,
 6x + 4y ≥ 12,
 x ≥ 0, y ≥ 0.

Answer 1

Hint: We find out the points through which the lines pass and draw a graph including all the line equations given in the conditions. Then we find a common point of these lines which gives the minimum value when substituted in z = 30x + 20y.

Complete step-by-step answer:
Given Data,
x + y ≤ 8, x + 2y ≥ 4, 6x + 4y ≥ 12, x ≥ 0, y ≥ 0.
Let us convert the given in-equations into equations w.r.t the conditions given for x and y.
⟹x + y = 8
Put x = 0, 1 and 2 in this equation, we get y = 8, 7, 6 respectively.
Hence we get the line x + y = 8 by joining these points (0, 8), (1, 7), (2, 6).
Hence the solution set of x + y ≤ 8 is the region in the xy plane which contains the origin.

⟹x + 2y = 4
Put x = 0, 2 and 4 in this equation, we get y = 2, 1, 0 respectively.
Hence we get this line by joining these points (0, 2), (2, 1), (4, 0).
Here clearly (0, 0) satisfies the in-equation x + 2y ≥ 4. Hence the region in the xy plane which does not contain the origin represents the solution set of this in-equation.

⟹6x + 4y = 12
Put x = 0, 1 and 2 in this equation, we get y =3, $\dfrac{3}{2}$ and 0 respectively.
Hence this line 6x+4y = 12 is formed by joining the points (0, 3), (1, $\dfrac{3}{2}$ ), (2, 0).
Here clearly (0, 0) satisfies the in-equation 6x + 4y ≥ 12. Hence the region in the xy plane which does not contain the origin represents the solution set of this in-equation.

Now we plot a graph which contains the line equations x + y = 8, x + 2y = 4 and 6x + 7y = 12 passing through their respective points. The solution set is the common part of all 3 equations, the shaded part of the graph:
It looks like

Now we are supposed to minimize z = 30x + 20y, i.e. find the minimum value of z w.r.t. a point from this shaded region of the graph.
As z = 30x + 20y, So we consider a few points with one coordinate zero, from the shaded region and find out which gives the minimum value of z.
The corner points of the shaded region are (0, 3), (4, 0), (8, 0) and (0, 8) and the point of intersection of the lines x+2y=4 and 6x+4y=12 i.e. (1, $\dfrac{3}{2}$ ) from the graph.
Or we could solve both the equations, x+2y=4 and 6x+4y=12 to find their point of intersection.
The value of z = 30x + 20y for (0, 3) = 0 + 60 = 60.
The value of z = 30x + 20y for (4, 0) = 120 + 0 = 120.
The value of z = 30x + 20y for (0, 8) = 0 + 160 = 160.
The value of z = 30x + 20y for (1, $\dfrac{3}{2}$ ) = 30 + 20× $\dfrac{3}{2}$ = 30 + 30 =60.
Therefore the minimum value of Z is 0 at point (1, $\dfrac{3}{2}$ ). Hence x = 1 and y = $\dfrac{3}{2}$ is the optimal solution of the given LPP.
Thus the optimal value of z = 30x + 20y is 0.

Note: In order to solve such types of problems, students must follow the graphical method of solution as done in the solution. For each of the lines given below, on the basis of the inequality given we find the region where the possible solution can exist, after combining the result of all the inequality, it is easier to find the solution. This problem can also be solved by solving the inequality together and then proceeding but the graphical method is quite easier.