Graphical Solution of Linear Inequalities of Two Variables
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Linear Inequalities and Linear Programming

88539 The minimum value for the \(L P P Z=6 x+2 y\), subject to \(2 x+y \geq 16, x \geq 6, y \geq 1\) is

1 34
2 47
3 44
4 24
Linear Inequalities and Linear Programming

88540 The optimal solution of the LPP. Maximum : \(Z=8 x+3 y\) subject to the constraints \(x+y \leq 3,4 x+y \leq 6, x \geq 0, y \geq 0\) is

1 \(x=0, y=0\)
2 \(x=1, y=2\)
3 \(x=0, y=3\)
4 \(x=\frac{3}{2}, y=0\)
MHT CET-2020
Linear Inequalities and Linear Programming

88541 The maximum value of \(Z=3 x+5 y\), subject to \(x+4 y \leq 24, y \leq 4, x \geq 0, y \geq 0\) is

1 120
2 72
3 44
4 20
MHT CET-2020
Linear Inequalities and Linear Programming

88542 The minimum value of \(Z=10 x+25 y\) subject to \(0 \leq \mathrm{x} \leq \mathbf{3 , 0} \leq \mathrm{y} \leq \mathbf{3}, \mathrm{x}+\mathrm{y} \geq \mathbf{5}\) is

1 95
2 80
3 105
4 30
[MHT CЕT-2019]
For More Updates join with Us
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Linear Inequalities and Linear Programming

88539 The minimum value for the \(L P P Z=6 x+2 y\), subject to \(2 x+y \geq 16, x \geq 6, y \geq 1\) is

1 34
2 47
3 44
4 24
Linear Inequalities and Linear Programming

88540 The optimal solution of the LPP. Maximum : \(Z=8 x+3 y\) subject to the constraints \(x+y \leq 3,4 x+y \leq 6, x \geq 0, y \geq 0\) is

1 \(x=0, y=0\)
2 \(x=1, y=2\)
3 \(x=0, y=3\)
4 \(x=\frac{3}{2}, y=0\)
MHT CET-2020
Linear Inequalities and Linear Programming

88541 The maximum value of \(Z=3 x+5 y\), subject to \(x+4 y \leq 24, y \leq 4, x \geq 0, y \geq 0\) is

1 120
2 72
3 44
4 20
MHT CET-2020
Linear Inequalities and Linear Programming

88542 The minimum value of \(Z=10 x+25 y\) subject to \(0 \leq \mathrm{x} \leq \mathbf{3 , 0} \leq \mathrm{y} \leq \mathbf{3}, \mathrm{x}+\mathrm{y} \geq \mathbf{5}\) is

1 95
2 80
3 105
4 30
[MHT CЕT-2019]
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Linear Inequalities and Linear Programming

88539 The minimum value for the \(L P P Z=6 x+2 y\), subject to \(2 x+y \geq 16, x \geq 6, y \geq 1\) is

1 34
2 47
3 44
4 24
Linear Inequalities and Linear Programming

88540 The optimal solution of the LPP. Maximum : \(Z=8 x+3 y\) subject to the constraints \(x+y \leq 3,4 x+y \leq 6, x \geq 0, y \geq 0\) is

1 \(x=0, y=0\)
2 \(x=1, y=2\)
3 \(x=0, y=3\)
4 \(x=\frac{3}{2}, y=0\)
MHT CET-2020
Linear Inequalities and Linear Programming

88541 The maximum value of \(Z=3 x+5 y\), subject to \(x+4 y \leq 24, y \leq 4, x \geq 0, y \geq 0\) is

1 120
2 72
3 44
4 20
MHT CET-2020
Linear Inequalities and Linear Programming

88542 The minimum value of \(Z=10 x+25 y\) subject to \(0 \leq \mathrm{x} \leq \mathbf{3 , 0} \leq \mathrm{y} \leq \mathbf{3}, \mathrm{x}+\mathrm{y} \geq \mathbf{5}\) is

1 95
2 80
3 105
4 30
[MHT CЕT-2019]
Join With Us for More Updates
Linear Inequalities and Linear Programming

88539 The minimum value for the \(L P P Z=6 x+2 y\), subject to \(2 x+y \geq 16, x \geq 6, y \geq 1\) is

1 34
2 47
3 44
4 24
Linear Inequalities and Linear Programming

88540 The optimal solution of the LPP. Maximum : \(Z=8 x+3 y\) subject to the constraints \(x+y \leq 3,4 x+y \leq 6, x \geq 0, y \geq 0\) is

1 \(x=0, y=0\)
2 \(x=1, y=2\)
3 \(x=0, y=3\)
4 \(x=\frac{3}{2}, y=0\)
MHT CET-2020
Linear Inequalities and Linear Programming

88541 The maximum value of \(Z=3 x+5 y\), subject to \(x+4 y \leq 24, y \leq 4, x \geq 0, y \geq 0\) is

1 120
2 72
3 44
4 20
MHT CET-2020
Linear Inequalities and Linear Programming

88542 The minimum value of \(Z=10 x+25 y\) subject to \(0 \leq \mathrm{x} \leq \mathbf{3 , 0} \leq \mathrm{y} \leq \mathbf{3}, \mathrm{x}+\mathrm{y} \geq \mathbf{5}\) is

1 95
2 80
3 105
4 30
[MHT CЕT-2019]
For More Updates join with Us