Maximize $$p = 15 x + 14 y$$ subject to the following. $$\begin{array} { l }
2 x + 2 y \leq 8 \
2 x + 8 y \leq 16 \
x \geq 0 , y \geq 0
\end{array}$$
A) $p = 60$ , $x = 4$ , $y = 4$
B) $p = 60$ , $x = 0$ , $y = 4$
C) $p = 60$ , $x = 4$ , $y = 0$
D) $p = 30$ , $x = 4$ , $y = 0$
E) $p = 30$ , $x = 0$ , $y = 4$

Question

Maximize $$p = 15 x + 14 y$$ subject to the following.   $$\begin{array} { l } 
2 x + 2 y \leq 8 \
2 x + 8 y \leq 16 \
x \geq 0 , y \geq 0
\end{array}$$  
A) $p = 60$ , $x = 4$ , $y = 4$
B)  $p = 60$ , $x = 0$ , $y = 4$
C)   $p = 60$ , $x = 4$ , $y = 0$
D)   $p = 30$ , $x = 4$ , $y = 0$
E)   $p = 30$ , $x = 0$ , $y = 4$

Quizplus · Accepted Answer

To solve this linear programming problem, we graph the constraints and find the feasible region. The vertices of the feasible region are the points where the constraints intersect. We then evaluate the objective function $p = 15x + 14y$ at each vertex to find the maximum value. - The first constraint $2x + 2y \leq 8$ simplifies to $x + y \leq 4$, which is a line that intersects the axes at $x = 4$ and $y = 4$.- The second constraint $2x + 8y \leq 16$ simplifies to $x + 4y \leq 8$, intersecting the axes at $x = 8$ and $y = 2$.- The non-negativity constraints $x \geq 0, y \geq 0$ limit the feasible region to the first quadrant.The vertices of the feasible region are at the intersections of these lines and the axes, which are $(0,0)$, $(4,0)$, and $(0,2)$. Evaluating the objective function at these points:- At $(0,0)$, $p = 15(0) + 14(0) = 0$.- At $(4,0)$, $p = 15(4) + 14(0) = 60$.- At $(0,2)$, $p = 15(0) + 14(2) = 28$.Therefore, the maximum value of $p$ is 60 when $x = 4$ and $y = 0$, which corresponds to choice C.

Maximize p=15x+14yp = 15 x + 14 yp=15x+14y Subject to the Following

Maximize $p = 15 x + 14 y$ Subject to the Following