Question
Maximize $Z = x + y$
Subject to
$-2\text{x}+\text{y}\leq1$
$\text{x}\leq2$
$\text{x}+\text{y}\leq3$
$\text{x},\text{y}\geq0$
Subject to
$-2\text{x}+\text{y}\leq1$
$\text{x}\leq2$
$\text{x}+\text{y}\leq3$
$\text{x},\text{y}\geq0$
✓
Answer
Converting the given inequations into equation,
-2x + y = 1, x = 2, x + y = 3, x= y = 0
Region represented by - 2x + y = 1:
Line - 2x + y = 1 meets coordinate axes at $\text{A}_1\Big(\frac{-1}{2},0\Big)$ and $B_1(0, 1)$, clearly, $(0, 0)$ satisfies $-2\text{x}+\text{y}\leq1$, so region containing origin represents $-2\text{x}+\text{y}\leq1$ in xy-plane.
Region represented by $\text{x}\leq2$:
Linex - 2 is parallel to y-axis and meets x-axis at $A_3(2, 0)$.
Clearly, (0, 0) satisfies $\text{x}\leq2$, so region containing origin represents $\text{x}\leq2$ in xy-plane.
Region represented by $\text{x}+\text{y}\leq3$:
Line $x + y - 3$ meets coordinate axes at $A_2(3, 0)$ and $B_2(0, 3)$.
Clearly, (0, 0) satisfies $\text{x}+\text{y}\leq3$, so region containing origin represents $\text{x}+\text{y}\leq3$ in xy-plane.
Region represented $\text{x},\text{y}\geq0$:
It represents first quadrant in xy-plane.
So, shaded region $OA_3PQ8$, represents the feasible region.
Coordinates of P(2, 1) is obtained by solving x + y = 3 and x = 2, $\text{Q}\Big(\frac{2}{3},\frac{7}{3}\Big)$ by solving -2x + y = 1 and x + y = 3.
The value of Z = x + y at
$\text{O}(0, 0) = 0 + 0 = 0$
$\text{A}_3(2, 0) = 2 + 0 = 2$
$\text{P}(2, 1) = 2 +1 = 2$
$\text{Q}\Big(\frac{2}{3},\frac{7}{3}\Big)=\frac{2}{3}+\frac{7}{3}=3$
$\text{B}_1(0, 1) = 0 + 1 = 1$
So, maximum Z = 3 is at every point on the line joining PQ.
Hence, maximum z = 3 at x = 2 and y = 1 Or $\text{x}=\frac{2}{3}$ and $\text{y}=\frac{7}{3}.$
-2x + y = 1, x = 2, x + y = 3, x= y = 0
Region represented by - 2x + y = 1:
Line - 2x + y = 1 meets coordinate axes at $\text{A}_1\Big(\frac{-1}{2},0\Big)$ and $B_1(0, 1)$, clearly, $(0, 0)$ satisfies $-2\text{x}+\text{y}\leq1$, so region containing origin represents $-2\text{x}+\text{y}\leq1$ in xy-plane.
Region represented by $\text{x}\leq2$:
Linex - 2 is parallel to y-axis and meets x-axis at $A_3(2, 0)$.
Clearly, (0, 0) satisfies $\text{x}\leq2$, so region containing origin represents $\text{x}\leq2$ in xy-plane.
Region represented by $\text{x}+\text{y}\leq3$:
Line $x + y - 3$ meets coordinate axes at $A_2(3, 0)$ and $B_2(0, 3)$.
Clearly, (0, 0) satisfies $\text{x}+\text{y}\leq3$, so region containing origin represents $\text{x}+\text{y}\leq3$ in xy-plane.
Region represented $\text{x},\text{y}\geq0$:
It represents first quadrant in xy-plane.
So, shaded region $OA_3PQ8$, represents the feasible region.
Coordinates of P(2, 1) is obtained by solving x + y = 3 and x = 2, $\text{Q}\Big(\frac{2}{3},\frac{7}{3}\Big)$ by solving -2x + y = 1 and x + y = 3.
The value of Z = x + y at
$\text{O}(0, 0) = 0 + 0 = 0$
$\text{A}_3(2, 0) = 2 + 0 = 2$
$\text{P}(2, 1) = 2 +1 = 2$
$\text{Q}\Big(\frac{2}{3},\frac{7}{3}\Big)=\frac{2}{3}+\frac{7}{3}=3$
$\text{B}_1(0, 1) = 0 + 1 = 1$
So, maximum Z = 3 is at every point on the line joining PQ.
Hence, maximum z = 3 at x = 2 and y = 1 Or $\text{x}=\frac{2}{3}$ and $\text{y}=\frac{7}{3}.$
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