The maximum value of $P=6x+8y$ subject to constraints $2x+y\le 30,\ x+2y\le 24$ and $x\ge 0,\ y\ge 0$ is
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Linear Programming  ETEA Model Test MCQS  Part  VIII
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By graphical method, the solution of linear programming problem Maximize $z=3{{x}_{1}}+5{{x}_{2}}$ Subject to $3{{x}_{1}}+2{{x}_{2}}\le 18$ , ${{x}_{1}}\le 4$ , ${{x}_{2}}\le 6$ , ${{x}_{1}}\ge 0$ , ${{x}_{2}}\ge 0$ is0${{x}_{1}}=2,\ {{x}_{2}}=0,\ z=6$0%0${{x}_{1}}=2,\ {{x}_{2}}=6,\ z=36$0%0${{x}_{1}}=4,\ {{x}_{2}}=3,\ z=27$0%0${{x}_{1}}=4,\ {{x}_{2}}=6,\ z=42$0%0
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The point at which the maximum value of $(x+y)$ subject to the constraints $2x+5y\le 100$ , $\frac{x}{25}+\frac{y}{49}\le 1$ , $x,\ y\ge 0$ is obtained, is0(10, 20)0%0(20, 10)0%0(15, 15)0%0$\left( \frac{50}{3},\ \frac{40}{3} \right)$0%0
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If $3{{x}_{1}}+5{{x}_{2}}\le 15$ , $5{{x}_{1}}+2{{x}_{2}}\le 10$ , ${{x}_{1}},\ {{x}_{2}}\ \ \ge 0$ then the maximum value of $5{{x}_{1}}+3{{x}_{2}}$ , by graphical method is0$12\frac{7}{19}$0%0$12\frac{1}{7}$0%0$12\frac{3}{5}$0%0120%0
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