Solve the linear programming problem: Maximize
$\begin{array}{c}
C=2 x+5 y \text { subject to } \
x+2 y \leq 10 \
3 x+2 y \leq 18 \
x \geq 0 \
y \geq 0
\end{array}$
A) 12
B) 23
C) 25
D) 32

Question

Solve the linear programming problem: Maximize 
$\begin{array}{c}
C=2 x+5 y \text { subject to } \
x+2 y \leq 10 \
3 x+2 y \leq 18 \
x \geq 0 \
y \geq 0
\end{array}$
A) 12
B) 23
C) 25
D) 32

Quizplus · Accepted Answer

The optimal solution to this linear programming problem can be found by evaluating the objective function $C = 2x + 5y$ at the vertices of the feasible region defined by the constraints. The constraints form a polygon in the first quadrant of the coordinate system. The vertices of this feasible region can be found by solving the system of equations formed by the intersection of the constraint lines:1. Intersection of $x + 2y = 10$ and $3x + 2y = 18$:   Solving these equations simultaneously gives $x = 4, y = 3$.2. Intersection of $x + 2y = 10$ and $x = 0$:   Setting $x = 0$ in the first constraint gives $y = 5$.3. Intersection of $3x + 2y = 18$ and $y = 0$:   Setting $y = 0$ in the second constraint gives $x = 6$.4. The points $(0, 0)$, $(0, 5)$, $(6, 0)$, and $(4, 3)$ are the vertices of the feasible region.Evaluating the objective function at these vertices:- At $(0, 0)$, $C = 2(0) + 5(0) = 0$.- At $(0, 5)$, $C = 2(0) + 5(5) = 25$.- At $(6, 0)$, $C = 2(6) + 5(0) = 12$.- At $(4, 3)$, $C = 2(4) + 5(3) = 8 + 15 = 23$.The maximum value of $C$ is 25, which occurs at $(0, 5)$.

Solve the Linear Programming Problem: Maximize C=2x+5y subject to x+2y≤103x+2y≤18x≥0y≥0\begin{array}{c}C=2 x+5 y \text { subject to } \\x+2 y \leq 10 \\3 x+2 y \leq 18 \\x \geq 0 \\y \geq 0\end{array}C=2x+5y subject to x+2y≤103x+2y≤18x≥0y≥0​

Solve the Linear Programming Problem: Maximize $\begin{array}{c}C=2 x+5 y \text { subject to } \\x+2 y \leq 10 \\3 x+2 y \leq 18 \\x \geq 0 \\y \geq 0\end{array}$