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

88552 If \(Z=10 x+25 y \quad\) subject to \(\mathbf{0} \leq \mathbf{x} \leq \mathbf{3}, \mathbf{0} \leq \mathbf{y} \leq \mathbf{3}, \mathbf{x}+\mathbf{y} \leq \mathbf{5}, \mathbf{x} \geq \mathbf{0}, \mathbf{y} \geq \mathbf{0}\) then \(\mathbf{Z}\) is maximum at the point
is maximum at the point

1 \((2,3)\)
2 \((2,4)\)
3 \((1,6)\)
4 \((4,3)\)
MHT CET-2020
Linear Inequalities and Linear Programming

88553 The L.P.P. to maximize \(z=x+y\) subject to \(x+y \leq 1,2 x+2 y \geq 6, x \geq 0, y \geq 0\) has

1 No solution
2 One solution
3 Two solutions
4 Infinite solutions
MHT CET-2020
Linear Inequalities and Linear Programming

88554 The objective function \(z=4 x_{1}+5 x_{2}\), subject to \(2 x_{1}+x_{2} \geq 7,2 x_{1}+3 x_{2} \leq 15, x_{2} \leq 3, x_{1}, x_{2} \geq 0\) has minimum value at the point

1 On X - axis
2 On Y - axis
3 At the origin
4 On the use parallel to \(x\)-axis
MHT CET-2017
Linear Inequalities and Linear Programming

88555 The shaded region is the solution set of the inequalities.

1 \(5 x+4 y \geq 20, x \leq 6, y \geq 3, x \geq 0, y \geq 0\)
2 \(5 x+4 y \leq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0\)
3 \(5 x+4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0\)
4 \(5 x+4 y \geq 20, x \geq 6, y \leq 3, x \geq 0, y \geq 0\)
Karnataka CET-2021
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Linear Inequalities and Linear Programming

88552 If \(Z=10 x+25 y \quad\) subject to \(\mathbf{0} \leq \mathbf{x} \leq \mathbf{3}, \mathbf{0} \leq \mathbf{y} \leq \mathbf{3}, \mathbf{x}+\mathbf{y} \leq \mathbf{5}, \mathbf{x} \geq \mathbf{0}, \mathbf{y} \geq \mathbf{0}\) then \(\mathbf{Z}\) is maximum at the point
is maximum at the point

1 \((2,3)\)
2 \((2,4)\)
3 \((1,6)\)
4 \((4,3)\)
MHT CET-2020
Linear Inequalities and Linear Programming

88553 The L.P.P. to maximize \(z=x+y\) subject to \(x+y \leq 1,2 x+2 y \geq 6, x \geq 0, y \geq 0\) has

1 No solution
2 One solution
3 Two solutions
4 Infinite solutions
MHT CET-2020
Linear Inequalities and Linear Programming

88554 The objective function \(z=4 x_{1}+5 x_{2}\), subject to \(2 x_{1}+x_{2} \geq 7,2 x_{1}+3 x_{2} \leq 15, x_{2} \leq 3, x_{1}, x_{2} \geq 0\) has minimum value at the point

1 On X - axis
2 On Y - axis
3 At the origin
4 On the use parallel to \(x\)-axis
MHT CET-2017
Linear Inequalities and Linear Programming

88555 The shaded region is the solution set of the inequalities.

1 \(5 x+4 y \geq 20, x \leq 6, y \geq 3, x \geq 0, y \geq 0\)
2 \(5 x+4 y \leq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0\)
3 \(5 x+4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0\)
4 \(5 x+4 y \geq 20, x \geq 6, y \leq 3, x \geq 0, y \geq 0\)
Karnataka CET-2021
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Linear Inequalities and Linear Programming

88552 If \(Z=10 x+25 y \quad\) subject to \(\mathbf{0} \leq \mathbf{x} \leq \mathbf{3}, \mathbf{0} \leq \mathbf{y} \leq \mathbf{3}, \mathbf{x}+\mathbf{y} \leq \mathbf{5}, \mathbf{x} \geq \mathbf{0}, \mathbf{y} \geq \mathbf{0}\) then \(\mathbf{Z}\) is maximum at the point
is maximum at the point

1 \((2,3)\)
2 \((2,4)\)
3 \((1,6)\)
4 \((4,3)\)
MHT CET-2020
Linear Inequalities and Linear Programming

88553 The L.P.P. to maximize \(z=x+y\) subject to \(x+y \leq 1,2 x+2 y \geq 6, x \geq 0, y \geq 0\) has

1 No solution
2 One solution
3 Two solutions
4 Infinite solutions
MHT CET-2020
Linear Inequalities and Linear Programming

88554 The objective function \(z=4 x_{1}+5 x_{2}\), subject to \(2 x_{1}+x_{2} \geq 7,2 x_{1}+3 x_{2} \leq 15, x_{2} \leq 3, x_{1}, x_{2} \geq 0\) has minimum value at the point

1 On X - axis
2 On Y - axis
3 At the origin
4 On the use parallel to \(x\)-axis
MHT CET-2017
Linear Inequalities and Linear Programming

88555 The shaded region is the solution set of the inequalities.

1 \(5 x+4 y \geq 20, x \leq 6, y \geq 3, x \geq 0, y \geq 0\)
2 \(5 x+4 y \leq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0\)
3 \(5 x+4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0\)
4 \(5 x+4 y \geq 20, x \geq 6, y \leq 3, x \geq 0, y \geq 0\)
Karnataka CET-2021
For More Updates join with Us
Linear Inequalities and Linear Programming

88552 If \(Z=10 x+25 y \quad\) subject to \(\mathbf{0} \leq \mathbf{x} \leq \mathbf{3}, \mathbf{0} \leq \mathbf{y} \leq \mathbf{3}, \mathbf{x}+\mathbf{y} \leq \mathbf{5}, \mathbf{x} \geq \mathbf{0}, \mathbf{y} \geq \mathbf{0}\) then \(\mathbf{Z}\) is maximum at the point
is maximum at the point

1 \((2,3)\)
2 \((2,4)\)
3 \((1,6)\)
4 \((4,3)\)
MHT CET-2020
Linear Inequalities and Linear Programming

88553 The L.P.P. to maximize \(z=x+y\) subject to \(x+y \leq 1,2 x+2 y \geq 6, x \geq 0, y \geq 0\) has

1 No solution
2 One solution
3 Two solutions
4 Infinite solutions
MHT CET-2020
Linear Inequalities and Linear Programming

88554 The objective function \(z=4 x_{1}+5 x_{2}\), subject to \(2 x_{1}+x_{2} \geq 7,2 x_{1}+3 x_{2} \leq 15, x_{2} \leq 3, x_{1}, x_{2} \geq 0\) has minimum value at the point

1 On X - axis
2 On Y - axis
3 At the origin
4 On the use parallel to \(x\)-axis
MHT CET-2017
Linear Inequalities and Linear Programming

88555 The shaded region is the solution set of the inequalities.

1 \(5 x+4 y \geq 20, x \leq 6, y \geq 3, x \geq 0, y \geq 0\)
2 \(5 x+4 y \leq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0\)
3 \(5 x+4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0\)
4 \(5 x+4 y \geq 20, x \geq 6, y \leq 3, x \geq 0, y \geq 0\)
Karnataka CET-2021