Question
Maximum $Z = 3x + 5y$
Subject to
$\text{x}+2\text{y}\leq20$
$\text{x}+\text{y}\leq15$
$\text{y}\leq5$
$\text{x},\text{y}\geq0$
Subject to
$\text{x}+2\text{y}\leq20$
$\text{x}+\text{y}\leq15$
$\text{y}\leq5$
$\text{x},\text{y}\geq0$
✓
Answer
Converting the given inequations into equation:-
$x + 2y = 20, x + y = 15, x = y = 0$
Region represented by $x + 2y = 20:$
Line x + 2y = 20 meets coordinate axes at $A_1(20, 0)$ and $B_1(0, 10)$, clearly, (0, 0) satisfies $\text{x}+2\text{y}\leq20$, so region containing origin represents $\text{x}+2\text{y}\leq20$ in xy -plane.
Region represented by $\text{x}+\text{y}\leq15$:
Line x + y = 15 meets coordinate axes at $A_2(15, 0)$ and $B_2(0, 15)$, clearly, (0, 0) satisfies $\text{x}+\text{y}\leq15$, so region containing origin represents x + y = 15 in xy-plane.
Region represented by $\text{y}\leq5$:
Line y = 5 is parallel to x-axis and meets at $B_3(0, 5)$ on y-axis.
Clearly (0, 0) satisfies $\text{y}\leq5$, so region containing origin represents y s 5 in xy-plane.
Region represented by $\text{x},\text{y}\geq0$:
It represent the first quadrant in xy-plane.
So, shaded region$ OA_2PB_3$ represents the feasible region.
Coordinate of P(10, 5) is obtained by solving $x + 2y = 20$ and $y - 5$
The value of $Z = 3x + 5y$ at
$O(0, 0) = 3(0) + 5(0) = 0$
$A_2(15, 0) = 3(15) + 5(0) = 45$
$P(10, 5) = 3(10) + 5(5) = 55$
$B_3(0, 5) = 3(0) + 5(5) = 25$
Hence, maximum $Z = 55$ at $x = 10$ and $y = 5$
$x + 2y = 20, x + y = 15, x = y = 0$
Region represented by $x + 2y = 20:$
Line x + 2y = 20 meets coordinate axes at $A_1(20, 0)$ and $B_1(0, 10)$, clearly, (0, 0) satisfies $\text{x}+2\text{y}\leq20$, so region containing origin represents $\text{x}+2\text{y}\leq20$ in xy -plane.
Region represented by $\text{x}+\text{y}\leq15$:
Line x + y = 15 meets coordinate axes at $A_2(15, 0)$ and $B_2(0, 15)$, clearly, (0, 0) satisfies $\text{x}+\text{y}\leq15$, so region containing origin represents x + y = 15 in xy-plane.
Region represented by $\text{y}\leq5$:
Line y = 5 is parallel to x-axis and meets at $B_3(0, 5)$ on y-axis.
Clearly (0, 0) satisfies $\text{y}\leq5$, so region containing origin represents y s 5 in xy-plane.
Region represented by $\text{x},\text{y}\geq0$:
It represent the first quadrant in xy-plane.
So, shaded region$ OA_2PB_3$ represents the feasible region.
Coordinate of P(10, 5) is obtained by solving $x + 2y = 20$ and $y - 5$
The value of $Z = 3x + 5y$ at
$O(0, 0) = 3(0) + 5(0) = 0$
$A_2(15, 0) = 3(15) + 5(0) = 45$
$P(10, 5) = 3(10) + 5(5) = 55$
$B_3(0, 5) = 3(0) + 5(5) = 25$
Hence, maximum $Z = 55$ at $x = 10$ and $y = 5$
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