The maximum value of Z = 5x + 2y,subject to t | vedclass

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Question

(A) To find the maximum value of $Z = 5x + 2y$,we identify the feasible region defined by the constraints:
$1$. $2x - y \geq 2$
$2$. $x + 2y \leq 8$

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) To find the maximum value of $Z = 5x + 2y$,we identify the feasible region defined by the constraints:
$1$. $2x - y \geq 2$
$2$. $x + 2y \leq 8$
$3$. $x, y \geq 0$
The boundary lines are $2x - y = 2$ and $x + 2y = 8$.
- For $2x - y = 2$,the intercepts are $(1, 0)$ and $(0, -2)$.
- For $x + 2y = 8$,the intercepts are $(8, 0)$ and $(0, 4)$.
To find the intersection point of $2x - y = 2$ and $x + 2y = 8$:
Multiply the first equation by $2$: $4x - 2y = 4$.
Adding this to $x + 2y = 8$,we get $5x = 12$,so $x = 2.4$.
Substituting $x = 2.4$ into $2x - y = 2$: $2(2.4) - y = 2 \implies 4.8 - y = 2 \implies y = 2.8$.
The vertices of the feasible region are $(1, 0)$,$(8, 0)$,and $(2.4, 2.8)$.
Now,evaluate $Z = 5x + 2y$ at these vertices:
- At $(1, 0)$: $Z = 5(1) + 2(0) = 5$
- At $(8, 0)$: $Z = 5(8) + 2(0) = 40$
- At $(2.4, 2.8)$: $Z = 5(2.4) + 2(2.8) = 12 + 5.6 = 17.6$
Comparing these values,the maximum value is $40$ at $(8, 0)$.

The maximum value of $Z = 5x + 2y$,subject to the constraints $2x - y \geq 2$,$x + 2y \leq 8$,and $x, y \geq 0$,is:

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