<p><strong>10. Maximise Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.</strong></p>

Ask & Answer Question

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

10. Maximise Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.

3 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

Maximise $z = x + y$ , subject to $x - y \leq - 1, - x + y \leq 0, x, y \geq 0$

The corresponding equation of the given inequalities are

$x - y = 1$ $\frac{x}{- 1} + \frac{y}{1} = 1$

$- x + y = 0$ $\Rightarrow$ $x = y$

$x, y = 0$ $x, y = 0$

The graph of the given inequalities is shown.

There is no common point in the two shaded region. Thus, there is no feasible region.

$∴$ Z has no maximum value.

0 Upvotes 0 Downvotes

Similar Questions for you

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 $a_{i j} = {\begin{matrix} J_{6 + i, 3} -_{i + 3, 3}, & i \leq j \\ 0, & i > j \end{matrix} . T h e n | a d j A^{- 1} | i s :$

alok kumar singh

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

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

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

$| A | = {(\frac{1}{32})}^{3} | B | = \frac{1}{105 . 2^{19}}$

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 :

Vishal Baghel

$α + β + γ = 2 π$

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

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

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

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

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

10. Maximise Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question