Download App

Linear programming — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSLinear programming1 Mark
MCQ
By graphical method, the solution of linear programming problem
Maximize $Z = 3x_1 + 5x_2$
Subject to
$3x_1 + 2x_2 \leq 18$
$x_1 \leq 4$
$x_2 \leq 6$
$x1 \geq 0, x2 \geq 0,$ is:
  • A
    $x_1 = 2, x_2 = 0, Z = 6$
  • B
    $x_1 = 2, x_2 = 6, Z = 36$
  • C
    $x_1 = 4, x_2 = 3, Z = 27$
  • D
    $x_1 = 4, x_2 = 6, Z = 42$
✓

Answer

We need to maximize the function $Z = 3x_4 + 5x_2$ First, we will convert the given inequations into equations, we obtain the following equations:
$3x_1 + 2x_2 = 18, x_1 = 4, x_2 = 6, x_1 = 0$ and $x_2 = 0$
Region represented by $3x_1 + 2x_2 \leq 18:$
The line $3x_1 + 2x_2 = 18$ meets the coordinate axes at $A(6, 0)$ and $B(0, 9)$ respectively.
By joining these points we obtain the line $3X1 + 2x2 = 18.$
Clearly $(0, 0)$ satisfies the inequation $3x_1 + 2x_2 = 18.$
So the region in the plane which contain the origin represents the solution set of the inequation $3x_1 + 2x_2 \leq 18.$
Region represented by $x_1 \leq 4:$
The line $x_1 = 4$ is the line that passes through $C(4, 0)$ and is parallel to the $Y$ axis.
The region to the left of the line $x_1 = 4$ will satisfy the inequation $x_1 \leq 4.$
Region represented by $x_2 \leq 6:$
The line $x_2 = 6$ is the line that passes through $D(0, 6)$ and is parallel to the $X$ axis.
The region below the line $x_2 = 6$ will satisfy the inequation $X_2 \leq 6.$
Region represented by $x_1 \geq 0$ and $x_2 \geq 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x_1 \geq 0$ and $x_2 \geq 0.$
The feasible region determined by the system of constraints, $3x_1 + 2x_2 \leq 18, x_1 \leq 4, x_2 \leq 6, x_1 \geq 0$ and $x_2 \geq 0$ are as follows

Corner points are $O(0, 0), D(0, 6), F(2, 6), E(4, 3)$ and $C(4, 0).$
The values of the objective function at these points are given in the following table.
Points
Value of $Z$
$O(0, 0)$ $3(0) + 5(0) = 0$
$D(0, 6)$ $3(0) + 5(6) = 30$
$F(2, 6)$ $3(2) + 5(6) = 36$
$E(4, 3)$ $3(4) + 5(3) = 27$
$C(4, 0)$ $3(4) + 5(0) = 12$
We see that the maximum value of the objective function $Z$ is 36 which is at $F(2, 6).$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Linear programming→Linear programming — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

If $f(x)=x^2 g(x)$ and $g(x)$ is twice differentiable, then $f^{\prime \prime}(x)$ is equal to
→
Difference between sample space and subset of sample space is considered as:
→
$\int\frac{\text{x}}{4+\text{x}^4}\text{ dx}$ is equal to:
  1. $\frac{1}{4}\tan^{-1}\text{x}^2+\text{C}$
  2. $\frac{1}{4}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)$
  3. $\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}^2}{2}\Big)$
  4. none of these.
→
If a line makes angle $\alpha,\beta$ and $\gamma$ with the axes respectively, then $\cos2\alpha+\cos2\beta+\cos2\gamma=$
→
The function $\text{f(x)}=|\cos\text{x}|$ is:
  1. Everywhere continuous and differentiable.
  2. Everywhere continuous but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
  3. Neither continuous nor differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
  4. None of these.
→
If A is any skew-symmetric matrix of odd order then ∣A∣ equals
  1. −1
  2. 0
  3. 1
  4. none of these
→
If $\int\frac{\cos8\text{x}+1}{\tan2\text{x}-\cot2\text{x}}\text{ dx}=\text{a}\cos8\text{x}+\text{C},$ then a =
  1. $-\frac{1}{16}$
  2. $\frac{1}{8}$
  3. $\frac{1}{16}$
  4. $-\frac{1}{8}$
→
The function $f(x)=\tan x-x$
→
If $\text{A}=\begin{bmatrix}1&\text{a}\\0&1\end{bmatrix},$ then $A^n ($where $n \in N)$ equals:
→
If in a binomial distribution $\text{n}=4,\text{P(X}=0)=\frac{16}{81},$ then $\text{P(X}=4)$ equals:
  1. $\frac{1}{16}$
  2. $\frac{1}{81}$
  3. $\frac{1}{27}$
  4. $\frac{1}{8}$
→