For a region defined by linear inequalities, a linear expression like $3x+2y$ reaches its maximum at a vertex (intersection of boundary lines).

One key vertex comes from solving the two equations $2x+y=11$ and $x+3y=12$ together.

Because the inequality is $3x+2y>k$ (not $\ge$), be careful about what happens when $k$ equals the maximum value.

Enter the first five inequalities:

• x+y<=7

• 2x+y<=11

• x+3y<=12

• x>=0

• y>=0

Desmos will shade the feasible region.

Graph the boundary lines as equalities (you can reuse the same expressions without the inequality signs). Click on intersection points that form the corners of the shaded region to read their coordinates (such as intersections with the axes and where two boundary lines cross).

For each vertex $(x,y)$ you found, evaluate 3x+2y (you can type it into Desmos and substitute the vertex coordinates or create points and then compute 3*A.x+2*A.y, etc.). аnіko.aі Find the greatest value among the vertices.

Create a slider by typing k=. Then graph 3x+2y>k and adjust $k$.

Observe the largest slider value where the shaded overlap still exists; because the inequality is strict (>), the system stops having solutions when $k$ reaches that maximum value or higher.

The first five inequalities define a feasible region in the $xy$-plane. To satisfy the last inequality $3x+2y>k$, the value of $k$ must be less than some value that $3x+2y$ can actually attain in that region.

© аniк​о‍.аi

So, find the maximum of $3x+2y$ on the feasible region. If $k$ is at least that maximum, the system will have no solution.

Because $3x+2y$ is linear and the region is formed by linear inequalities, the maximum occurs at a vertex.

Relevant vertices come from intersections of boundary lines and the axes:

• From $x=0$ and $x+3y=12$: $(0,4)$.
• From $y=0$ and $2x+y=11$: $x=\frac{11}{2}$, so $\left(\frac{11}{2},0\right)$.
• The origin $(0,0)$.

A key vertex comes from intersecting $2x+y=11$ and $x+3y=12$:

Multiply the first equation by $3$ and subtract the second:

Then $y=11-2x=11-\frac{42}{5}=\frac{13}{5}$, giving the vertex $\left(\frac{21}{5},\frac{13}{5}\right)$.

(Each of these points satisfies $x+y\le 7$, so they are feasible.)

Compute $3x+2y$:

• At $(0,0)$: $3(0)+2(0)=0$.
• At $(0,4)$: $3(0)+2(4)=8$.
• At $\left(\frac{11}{2},0\right)$: $3\left(\frac{11}{2}\right)+2(0)=\frac{33}{2}$.
• At $\left(\frac{21}{5},\frac{13}{5}\right)$:

So the maximum value of $3x+2y$ on the feasible region is $\frac{89}{5}$.

The system requires $3x+2y>k$.

• If $k$ is at least the maximum value of $3x+2y$ on the feasible region, then $3x+2y>k$ cannot happen anywhere in the region, so there is no solution.
• If $k$ is less than that maximum value, then at least one feasible point has $3x+2y>k$, so a solution exists.

Since the maximum of $3x+2y$ is $\frac{89}{5}$, the system has no solution when $k \ge \frac{89}{5}$.