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

3. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0

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Vishal Baghel | Contributor-Level 10

4 months ago

Maximise $z = 5 x + 3 y$

Subject to $2 x + 5 y \leq 15, 5 x + 2 y \leq 10, x \geq 0, y \geq 0$

The corresponding equation of the above linear inequalities are

$\begin{array}{l} 3 x + 5 y = 15 \\ 5 x + 2 y = 10 & x = 0, y = 0 \end{array}$

$\Rightarrow \frac{x}{5} + \frac{y}{3} = 1$

$\begin{array}{l} \frac{x}{2} + \frac{y}{5} = 1 \\ x = 0, y = 0 \end{array}$

The graph of its given inequalities.

The shaded region OABC is the feasible region which is bounded with the corner points

$O (0, 0), A (2, 0), B (\frac{20}{19}, \frac{45}{19}) & C (0, 3)$

The values of Z at these points are

Therefore the maximum value of Z is $\frac{235}{19} a t, B (\frac{20}{19}, \frac{45}{19})$

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The system of linear equations

3x -2y -kz = 10

2x -4y -2z = 6

x + 2y -z = 5m

is inconsistent if

alok kumar singh

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

= -8 (-3 + k)

For inconsistent $Δ = 0 \Rightarrow k = 3$

$Δ_{x} = | \begin{matrix} 10 & - 2 & - k \\ 6 & - 4 & - 2 \\ 5 m & 2 & - 1 \end{matrix} | = 32 - 40 m \neq 0 \Rightarrow m \neq \frac{4}{5}$

If [x] denotes the greatest integer less than or equal to x, then the value of the integral $\int_{- π / 2}^{π / 2} [[x] - s i n x] d x$ is equal to:

Vishal Baghel

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

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

by using property $(\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x)$

Adding (i) and (ii) we get 2l = $\int_{- \frac{π}{2}}^{\frac{π}{2}} (- 2) d x = - 2 π \Rightarrow l = - π$

Let $J_{n, m} = \int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m} - 1} d x, \forall n > m$ and n, $m \in N$ . Consider a matrix $A = {[a_{i j}]}_{3 \times 3}$ where $a_{i j} = {\begin{matrix} J_{6 + i, 3} -_{i + 3, 3}, & i \leq j \\ 0, & i > j \end{matrix} . T h e n | a d j A^{- 1} | i s :$

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$l_{n, m} = \int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m} - 1} d x, m, n \in N, n > m$

$l_{6 + i, 3} - l_{3 + i, 3}$

$A = \frac{1}{2^{5}} [\begin{matrix} \frac{1}{5} & \frac{1}{5} & \frac{1}{5} \\ 0 & \frac{1}{12} & \frac{1}{12} \\ 0 & 0 & \frac{1}{28} \end{matrix}]$ $= \frac{B}{32}$

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

$x + (c o s γ) y + (c o s β) z = 0$

$(c o s γ) x + y + (c o s α) z = 0$

$(c o s β) x + (c o s α) y + z = 0$

has :

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$α + β + γ = 2 π$

$Δ = | \begin{matrix} 1 & c o s γ & c o s β \\ c o s γ & 1 & c o s α \\ c o s β & c o s α & 1 \end{matrix} |$

$= 1 - c o s^{2} α - c o s^{2} β - c o s^{2} γ + 2 c o s α . c o s β . c o s γ$

$= - c o s γ c o s (α - β) + c o s γ c o s (α - β) = 0$

If the system of linear equations

2x + y – z = 3

x – y – z = a

3x + 3y + bz = 3

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

Vishal Baghel

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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3. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0

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