Question

Use graphical methods to solve this linear programming problem.

Maximize z = 3x + 4y
subject to 2x - 3y ≤ 12
x + y ≥ 3
3x + 4y ≥ 24
x ≥ 0
y ≥ 0

What is the maximum value of​ z? Select the correct choice​ below, and if​ necessary, fill in the answer box to complete your choice.

A. The maximum value is ___ (Simplify answer)
B. There is no maximum.

At what​ point(s) does the maximum value of z​ occur? Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A. The maximum value of z occurs only at the​ point(s) ___
​(Type an ordered pair. Use a comma to separate answers as​ needed.)
B. The maximum value of z occurs at the points ___ and at all points on the line segment connecting them. ​(Type an ordered pair. Use a comma to separate answers as​ needed.)
C. There is no maximum value of z.

Answer 1

B. There is no maximum value.

C. There is no maximum value of z.

Explanation:

To solve this linear programming problem using graphical methods, start by graphing the feasible region determined by the given constraints:

• 2x - 3y ≤ 12
• x + y ≥ 3
• 3x + 4y ≥ 24
• x ≥ 0
• y ≥ 0

We can graph each of these inequalities by first plotting the corresponding boundary line, and then shading in the appropriate region.

Rearrange the first three inequalities to isolate y:

`\boxed{\begin{aligned}2x - 3y & \leq 12\\2x - 12 & \leq 3y \\3y &\geq 2x - 12\\y& \geq (2)/(3)x-4\end{aligned}}`
`\boxed{\begin{aligned}x + y & \geq 3\\y & \geq -x+3\\ \phantom{w}\\\phantom{\frac12}\end{aligned}}`
`\boxed{\begin{aligned}3x + 4y & \geq 24\\4y & \geq -3x + 24\\y & \geq -(3)/(4)x + 6\\\phantom{w}\end{aligned}}`

Graph the inequalities.

• If the inequality sign is ≥, draw a solid line and shade above the line.
• If the inequality sign is ≤, draw a solid line and shade below the line.

The feasible region is the region that is shaded by all of the inequalities.

Please see the attached graph.

A bounded feasible region may be enclosed in a circle and will have both a maximum value and a minimum value for the objective function.

If a feasible region is unbounded, and the coefficients on the objective function are all positive, then an unbounded feasible region will have a minimum but no maximum, since there is no limit on how big it can get.

Therefore, as the feasible region for the given constraints is unbounded, and the coefficients of the objective function z = 3x + 4y are all positive, there is no maximum value.