50 questions · timed · auto-graded
Solution:
4a + 0b = 2a + 4b
4a = 2a + 4b
4a - 2a = 4b
2a = 4b
a = 2b
Solution:
4a + 0b = 2a + 4b
4a = 2a + 4b
4a - 2a = 4b
2a = 4b
a = 2b
Solution:
4a + 0b = 2a + 4b
4a = 2a + 4b
4a - 2a = 4b
2a = 4b
a = 2b
Solution:
| Corner points | Corresponding value of Z = 3x - 4y |
| (0, 0) (5, 0) (6, 5) (6, 8) (4, 10) (0, 8) | 0 15 (Maxmimum) -2 -14 -28 -32 |
Hence, maximum of Z occurs at (5, 0) and its maximum value is 27.
Solution:
Better approch is with graphs.Considering graphs in eqaution we get
$\text{x}^{2}-\text{x}\sin\text{x}-\cos\text{x}=0$
$\text{x}^{2}=\text{x}\sin\text{x}+\cos\text{x}$
Let $\text{f}(\text{x})=\text{x}^{2},\text{g}(\text{x})=\text{x}\sin\text{x}+\cos\text{x}$
Using graphical methods,we can do the graph of f(x) and g(x)
The graph f(x) and g(x) intersects at two points between $(-\infty,\infty)$
Solution:
The feasibe solution of a inear programming probem (LP) is a solution that must satisfy all of the problems constraints simultaniously.
| Column A | Column B |
| Maximum of Z | 325 |
Solution:
Linear programming problem may be defined as the problem of maximizing or minimizing a linear function subject to linear constraints.
The objective function in a linear program is a maximization function.
| | Number of cars manufactured | ||
| Colour | Vento | Creta | Wagonr |
| Red | 65 | 88 | 93 |
| White | 54 | 42 | 80 |
| Black | 66 | 52 | 88 |
| Sliver | 37 | 49 | 74 |
Solution:
The number of silver Vento car = 37 (from the table)
Twice the number of silver Vento cars = 2 × 37 = 74
Now from table we can see that silver WagonR is only car type having 74 cars
Solution:
There are different methods to solve an linear programming problem.
Such as Graphical method, Simplex method, Ellipsoid method, Interior point methods.
Solution:
The problem associated with LLP is single objective.
Solution:
| Corner points | Corresponding value of Z = 3x - 4y |
| (0, 0) (5, 0) (6, 5) (6, 8) (4, 10) (0, 8) | 0 15 (Maximum) -2 -14 -28 -32 (Minimum) |
Here, maximum value of Z + minimum value of Z = 15 - 32 = -17.
Solution:
In linear programming, objective function and objective constraints are linear.
Any linear programming problem must have the following properties:-
The relationship between variables and constraints must be linear.
The constraints must be non-negative.
objective function must be linear.
Solution:
The objective of Linear Programming Problems (LPP) is to minimize or maximize the function.
Solution:
In linear programming, transportation model are applied to problems related to the study of efficient transportation routes.
For oil companies, how effectively the available resources are transported to different destinations with minimum cost.
Solution:
We know that, for a cartesian polygon, the maximum value occurs at the corner points or vertices of the polygon.
Given z = 300x + 190y
By substituting A(0, 0) in the equation we get z = 0
By substituting B(16, 0) in the equation we get z = 4800
By substituting C(8,16) in the equation we get z = 5440
By substituting D(0, 24) in the equation we get z = 4560
Hence the minimum value of Z occured at C(0, 0) with z = 0
Solution:
Given, a = b Multiplying both sides by x.
ax = bx.
Solution:
$\text{x}\leq4$ and $\text{x}\leq-4$
$\Rightarrow\text{x}\leq-4$
Also, $\text{y}\geq0$ and $\text{y}\leq0$
$\Rightarrow\text{y}=0$
Hence the solutione is $\text{x}\leq-4,\text{y}=0.$
Solution:
Optimal solution of LPP has three types.
Hence, it has infinite solution if it admits two optimal solution.
Solution:
Let z0 be the maximum value of z in the feasible region.
Since maximum occurs at both (15, 15) and (0, 20)$, the value z0 is attained at both (15, 15) and (0, 20).
⟹ z0 = p(15) + q(15) and z0 = p(0) + q(20)
⟹ p(15) + q(15) = p(0) + q(20)
⟹ 15p = 5q
⟹ 3p = q
Solution:
Given: d1 = 2cm
d2 = 4cm
Since the diameter are 2cm and 4cm.
The replacement ratio of the two pipes are 1cm and 2cmr1 = 1cm
r2 = 2cm
Square of the ratio of the pipes are 1 and 4
$\therefore$ The ratio of rates of flow in two pipes $=1:\frac{1}{4}$
$\Rightarrow\frac{1}{4}$
Solution:
| Corner points | Corresponding value of F = 4x + 6y |
| (0, 2) | 12 (Minimum) |
| (3, 0) | 12 (Minimum) |
| (6, 0) | 24 |
| (6, 8) | 72 (Maxmimum) |
| (0, 5) | 30 |
Solution:
We check the value of the z at each of the corner points.
A (0, 5) -z = -50x + 20y = -50(0) + 20(5) = 100
At B (0, 3) -z = -50x + 20y = -50(0) + 20(3) = 60
At C (1, 0) -z = -50x + 20y = -50(1) + 20(0) = -50
At D (6, 0) -z = -50x + 20y = -50(6) + 20(0) = -300
Hence, we see that z is minimum at D(6, 0) and minimum value is -300.
Solution:
For linear programming, the constraints must be linear.
Solution:
Given,
A = {1, 2, 3}
B = {3, 4, 5}
C = {4, 6}
Now, $\text{B}\cap\text{C}=\{{4\}}$
$\therefore\text{A}\times(\text{B}\cap\text{C})=\{(1,4),(2,4),(3,4)\}$
Solution:
It is known that the optimal value of the objective function is attained at any of the corner point.
Thus, the potimal value of the objective function is attined at the points given by corner points of the feasible region.
| | Number of cars manufactured | ||
| Colour | Vento | Creta | WagonR |
| Red | 65 | 88 | 93 |
| White | 54 | 42 | 80 |
| Black | 66 | 52 | 88 |
| Sliver | 37 | 49 | 74 |
Solution:
The number of Black cars manufactured.
= no. of black V ento + no. of black Creta + no. of black W agon R.
= 66 + 52 + 88 = 206
Solution:
The graph of the linear equation is a straight line.
If the terminal points are connected by a straight line then the given constraints are linear equations which may include inequalities.
Solution:
The feasible region as shown in the figure, has objective function F = 3x - 4y
| Corner points | Corresponding value of Z = 3x - 4y |
| (0, 0) (12, 6) (0, 4) | 0 12 (maximum) -16 (minimum) |
Hence, the maximum value of F is 12.
Solution:
Maximum of Z occurs at (25, 20) and at (0, 30).
Hence, equating the vales of Z at these points, we get 25p + 20q = 30q
$\therefore$ 5p = 2q
This is the required relation.
Also as p, q > 0, the value of Z is always positive and hence, is greater at (25, 20) and at (0, 30) than at (0,10) and (5, 5).
Solution:
given, n2 > 10 and n > 0 multiplying both equations we get n3 > 0
so, it may be greater than or less than 50.
Hence, quantity A is may be greater or smaller than B
Solution:
A sensitivity analysis is performed to determine the sensitivity of the solution to changes in parameters.
Solution:
Let A = Part X is not defective
Probability of A is $\text{P}(\text{A})=\frac{91}{100}$
B = Part Y is not defective.
Probability of B is $\text{P}(\text{B})=\frac{95}{100}$
Required probability
$=\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})\text{P}(\text{B})=\frac{91}{100}\times\frac{95}{100}=\frac{8645}{10000}$
Solution:
$28+7\text{z}>34+5.50\text{z}$
$\rightarrow1.50\text{z}>6$
$\rightarrow\text{z}>\frac{6}{1.5}\text{z}>4$
Solution:
We need to maximize the function
Z = x + y
Converting the given inequations into equations, we obtain x + 2y = 70, 2x + y = 95, x = 0 and y = 0
Region represented by x + 2y ≤ 70:
The line x + 2y = 70 meets the coordinate axes at A(70, 0) and B(0, 35) respectively.
By joining these points we obtain the line x + 2y = 70.
Clearly (0, 0) satisfies the inequation x + 2y ≤ 70.
So, the region containing the origin represents the solution set of the inequation x + 2y ≤ 70.
Region represented by 2x + y ≤ 95:
The line 2x + y = 95 meets the coordinate axes at $\text{C}\Big(\frac{95}{2},0\Big)$ and D(0, 95) respectively.
By joining these points we obtain the line 2x + y = 95.
Clearly (0, 0) satisfies the inequation 2x + y ≤ 95.
So, the region containing the origin represents the solution set of the inequation 2x + y ≤ 95.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, and y ≥ 0, are as follows.
The corner points of the feasible region are O(0, 0), $\text{C}\Big(\frac{95}{2},0\Big)$, E(40, 15) and B(0, 35).
The values of Z at these corner points are as follows.
| $\text{Corner point}$ | $\text{Z} = \text{x} + \text{y}$ |
| $\text{O}(0, 0)$ | $0 + 0 = 0$ |
| $\text{C}\Big(\frac{95}{2},0\Big)$ | $\frac{95}{2}+0,2=\frac{95}{2}$ |
| $\text{E}(40, 1)$ | $40 + 15 = 55$ |
| $\text{B}(0, 35)$ | $0 + 35 = 35$ |
We see that the maximum value of the objective function Z is 55 which is at (40, 15).
Solution:
If there is no point in the feasible set, there is no feasible solution of the linear programming model.
In linear programming, lack of points for a solution set is said to have no feasible solution.
Solution:
We know that $ \text{A}.\text{M}.\geq\text{G}.\text{M}.$
Therefore, $ \frac{{\text{b}^2+\text{c}^2}}{2}\geq\sqrt{\text{b}^2\text{c}^2}$
$\Rightarrow\text{b}^{2}+\text{c}^{2}\geq2\text{bc}$
Multiplying a on both sides doesn’t change the inequality.
Since, given that a is positive.
$\Rightarrow\text{a}(\text{b}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(1)$
Similarly, $\text{b}(\text{a}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(2)$
and $\text{c}(\text{a}^{2}+\text{c}^{2})\geq2\text{abc}\ ...(3)$
adding (1), (2) and (3) we get
$\text{a}(\text{b}^{2}+\text{c}^{2})+\text{b}(\text{a}^{2}+\text{c}^{2})+\text{c}(\text{a}^{2}+\text{b}^{2})\geq2\text{abc}+2\text{abc}+2\text{abc}$
$\Rightarrow\text{a}(\text{b}^{2}+\text{c}^{2})+\text{b}(\text{a}^{2}+\text{b}^{2})+\text{c}(\text{a}^{2}+\text{b}^{2})\geq6\text{abc}$
Therefore $\lambda$ is 6
Solution:
By putting the value of point (2, 3) in 2x + 3y - 12, we get;
2(2) + 3(3) = -12
= 4 + 9 - 12
= 13 - 12
= 1 which is greater than 0.
Solution:
The end points of the figure which forms as per the given condition are (0, 0), (3, 0), (0, 3), (3, 2), (2, 3) We check the value of z at these points.
At (0, 3), z = 0 + 75 = 75 At (3, 0), z = 30 + 0 = 30 At (0, 0), z = 0 At (3, 2), z = 30 + 50 = 80 At (2, 3), z = 20 + 75 = 95
Therefore, the maximum value of z turns out to be 95.