Given,

Z = 4x + 2y

Subjected to constraints,

2x + 3y ≤ 18

x + y ≥ 10

x ≥ 0

y ≥ 0

Consider, the inequalities as equalities for some time,

2x + 3y = 18 and x + y = 10,

If we convert these into intercept line format equations, we get,

[Dividing the whole equation with the right hand side number of the equation]

and

and

From this form of line, we can say that the line 2x + 3y = 18 meets the x-axis at (9,0) and y-axis at (0,6).

This shows the inequality 2x + 3y ≤ 18 holds good in the below blue colored region.

Similarly from the intercept line format, we can say that the line x + y = 10 meets the x-axis at (10,0) and y-axis at (0,10).

This shows the inequality x + y ≥ 10 holds above in the green colored region.

Now considering the inequalities, x ≥ 0 and y ≥ 0, this clearly shows the region where both x and y are positive. This represents the 1^st quadrant of the graph.

So, the solutions of the LPP are in the first quadrant where the inequalities meet.

Now by plotting both graphs 2x + 3y ≤ 18 and x + y ≥ 10 we get the below graph.

We can clearly see that, there is no area in the 1^st quadrant where the two inequalities met.

This clearly says that there is no solution for the LPP with the given constraints.

Hence the option D, is the solution to the problem.