<p><strong>2. Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.</strong></p>

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Ncert Solutions Maths class 12th Maths Class 12th Linear Programming

2. Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

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Vishal Baghel | Contributor-Level 10

4 months ago

Minimize $z = - 3 x + 4 y$

Subject to $x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0, y \geq 0$

The corresponding equation of the given inequalities are

$\begin{array}{l} x + 2 y = 8 \\ 3 x + 2 y = 12 \\ x = 0, y = 0 \end{array}$

$\Rightarrow \frac{x}{8} + \frac{y}{4} = 1$

$\begin{array}{l} \frac{x}{4} + \frac{y}{6} = 1 \\ x = 0, y = 0 \end{array}$

The graph is shown below.

The bounded region OABC is the feasible region with the corner points O (0,0), A (4,0), B (2,3), and C (0,4

The value of Z at these points are

Therefore, the minimum value of Z is -12 at (4,0).

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The system of linear equations

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$Δ - | \begin{matrix} 3 & - 2 & - k \\ 2 & - 4 & - 2 \\ 1 & 2 & - 1 \end{matrix} | = | \begin{matrix} 3 & - 2 & - k \\ 0 & - 8 & 0 \\ 1 & 2 & - 1 \end{matrix} |, [R_{2} - R_{1} - 2 R_{3}]$

= -8 (-3 + k)

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$l = \int_{- \frac{π}{2}}^{\frac{π}{2}} ([x] + [- s i n x]) d x$ . (i)

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If α + β + = 2π, then the system of equations

$x + (c o s γ) y + (c o s β) z = 0$

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$α + β + γ = 2 π$

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If the system of linear equations

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x – y – z = a

3x + 3y + bz = 3

has infinitely many solution, then α + β - αβ is equal to__________

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Given 2x + y – z = 3 . (i)

x – y – z = α . (ii)

3x + 3y + βz = 3 . (iii)

(i) x 2 – (ii) – (iii) – (1 + β) z = 3 - α

For infinite solution 1 + β = 0 = 3 - α

=> α = 3, β = -1

So, α + β - αβ = 5

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2. Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

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