Question
Maximum $Z = 10x + 6y$
Subject to
$3\text{x}+\text{y}\leq12$
$2\text{x}+5\text{y}\leq34$
$\text{x},\text{y}\geq0$
Subject to
$3\text{x}+\text{y}\leq12$
$2\text{x}+5\text{y}\leq34$
$\text{x},\text{y}\geq0$
✓
Answer
Converting the given inequations in to equations.
$3x + y = 12, 2x + 5y = 34, x = y = 0$
Region represented by $3\text{x}+\text{y}\leq12$:
Line $3x + y = 12$ meets the coordinate axes at $A_1(4, 0)$ and $B(0, 12)$, clearly, $(0, 0)$ satisfies $3\text{x}+\text{y}\leq12$, so, region containing origin is represented by $3\text{x}+\text{y}\leq12$ in xy-plane.
Region represented by $2\text{x}+5\text{y}\leq34$:
Line $2x +y = 34$ meets coordinate axes at $A_2$_$(17, 0)$ and $\text{B}\Big(0,\frac{34}{5}\Big)$ clearly, $(0, 0)$ satisfies the $2\text{x}+5\text{y}\leq34$ so, region containing origin represents $2\text{x}+5\text{y}\leq34$ in xy-plane.
Region represented by $\text{x},\text{y}\geq0$:
It represent the first quadrant in xy-plane
Therefore, shaded area $OA_1PB_2$ is the feasible region.
The coordinate of $P(2, 6)$ is obtained by solving $2x + 5y = 34$ and $3x + y = 12$
The value of $Z = 10x + 6y at$
$\text{O}(0, 0) = 10(0) + 6(0) = 0$
$\text{A}_1(4, 0) = 10(4) + 6(0) = 40$
$\text{P}(2, 6) = 10(2) + 6(6) = 56$
$\text{B}_2\Big(0,\frac{34}{5}\Big)=10(0)+6\Big(\frac{34}{5}\Big)=\frac{204}{5}=40\frac{4}{5}$
Hence, maximum $Z = 56 at x = 2, y = 6.$
$3x + y = 12, 2x + 5y = 34, x = y = 0$
Region represented by $3\text{x}+\text{y}\leq12$:
Line $3x + y = 12$ meets the coordinate axes at $A_1(4, 0)$ and $B(0, 12)$, clearly, $(0, 0)$ satisfies $3\text{x}+\text{y}\leq12$, so, region containing origin is represented by $3\text{x}+\text{y}\leq12$ in xy-plane.
Region represented by $2\text{x}+5\text{y}\leq34$:
Line $2x +y = 34$ meets coordinate axes at $A_2$_$(17, 0)$ and $\text{B}\Big(0,\frac{34}{5}\Big)$ clearly, $(0, 0)$ satisfies the $2\text{x}+5\text{y}\leq34$ so, region containing origin represents $2\text{x}+5\text{y}\leq34$ in xy-plane.
Region represented by $\text{x},\text{y}\geq0$:
It represent the first quadrant in xy-plane
Therefore, shaded area $OA_1PB_2$ is the feasible region.
The coordinate of $P(2, 6)$ is obtained by solving $2x + 5y = 34$ and $3x + y = 12$
The value of $Z = 10x + 6y at$
$\text{O}(0, 0) = 10(0) + 6(0) = 0$
$\text{A}_1(4, 0) = 10(4) + 6(0) = 40$
$\text{P}(2, 6) = 10(2) + 6(6) = 56$
$\text{B}_2\Big(0,\frac{34}{5}\Big)=10(0)+6\Big(\frac{34}{5}\Big)=\frac{204}{5}=40\frac{4}{5}$
Hence, maximum $Z = 56 at x = 2, y = 6.$
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