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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