The difference between the maximum and minimu | vedclass

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(A) The feasible region is determined by the constraints $x + 3y \leqslant 60$,$x + y \geqslant 10$,$x - y \leqslant 0$,$x \geqslant 0$,and $y \geqsla

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) The feasible region is determined by the constraints $x + 3y \leqslant 60$,$x + y \geqslant 10$,$x - y \leqslant 0$,$x \geqslant 0$,and $y \geqslant 0$.
First,we find the vertices of the feasible region by solving the intersection points of the boundary lines:
$1$. Intersection of $x - y = 0$ and $x + y = 10$: $2x = 10 \implies x = 5, y = 5$. Vertex: $(5, 5)$.
$2$. Intersection of $x - y = 0$ and $x + 3y = 60$: $4y = 60 \implies y = 15, x = 15$. Vertex: $(15, 15)$.
$3$. Intersection of $x + y = 10$ and $x = 0$: $y = 10$. Vertex: $(0, 10)$.
$4$. Intersection of $x + 3y = 60$ and $x = 0$: $3y = 60 \implies y = 20$. Vertex: $(0, 20)$.
Now,we evaluate $Z = 3x + 5y$ at each vertex:
- At $(5, 5)$: $Z = 3(5) + 5(5) = 15 + 25 = 40$.
- At $(15, 15)$: $Z = 3(15) + 5(15) = 45 + 75 = 120$.
- At $(0, 10)$: $Z = 3(0) + 5(10) = 50$.
- At $(0, 20)$: $Z = 3(0) + 5(20) = 100$.
The maximum value is $120$ and the minimum value is $40$.
The difference is $120 - 40 = 80$.

The difference between the maximum and minimum values of the objective function $Z = 3x + 5y$,subject to the constraints $x + 3y \leqslant 60$,$x + y \geqslant 10$,$x - y \leqslant 0$,$x \geqslant 0$,$y \geqslant 0$ is

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