Tag: structure of linear programming model

Questions Related to structure of linear programming model

Let $a _1,a _2....,a _n$ be a non negative real number such that $a _1+a _2....+a _n=m$ and let $S=\underset{i<j}\sum a _ia _j$, then

business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. $S\leq \dfrac {m^2}2$

  2. $S> \dfrac {m^2}4$

  3. $S< \dfrac {m}2$

  4. $S> \dfrac {m^2}2$

Show answer
Correct Option: A

A firm manufactures three products $A,B$ and $C$. Time to manufacture product $A$ is twice that for $B$ and thrice that for $C$ and if the entire labour is engaged in making product $A,1600$ units of this product can be produced.These products are to be produced in the ratio $3:4:5.$ There is demand for at least $300,250$ and $200$ units of products $A,B$ and $C$ and the profit earned per unit is Rs.$90,$ Rs$40$ and Rs.$30$ respectively.

Rawmaterial Requirement per unit product(Kg)A Requirement per unit product(Kg)B Requirement per unit product(Kg)C Total availability (kg)
$P$ $6$ $5$ $2$ $5,000$
$Q$ $4$ $7$ $3$ $6,000$

Formulate the problem as a linear programming problem and find all the constraints for the above product mix problem.

business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. $3{x} _{1}-4{x} _{2}=0$ and $5{x} _{2}-4{x} _{3}=0$ where ${x} _{1},{x} _{2},{x} _{3}\ge0$

  2. $4{x} _{1}-3{x} _{2}=0$ and $5{x} _{2}-4{x} _{3}=0$ where ${x} _{1},{x} _{2},{x} _{3}\ge0$

  3. $4{x} _{1}-3{x} _{2}=0$ and $4{x} _{2}-5{x} _{3}=0$ where ${x} _{1},{x} _{2},{x} _{3}\ge0$

  4. $4{x} _{1}-3{x} _{2}=0$ and $5{x} _{2}-4{x} _{3}=0$ where ${x} _{1},{x} _{2},{x} _{3}\le0$

Show answer
Correct Option: B
Explanation:

Formulation of L.P Model
Let ${x} _{1},{x} _{2}$ and ${x} _{3}$ denote the number of units of products $A,B$ and $C$ to be manufactured .
Objective is to maximize the profit.
i.e., maximize $Z=90{x} _{1}+40{x} _{2}+30{x} _{3}$
Constraints can be formulated as follows:
For raw material $P, 6{x} _{1}+5{x} _{2}+2{x} _{3}\le5,000$
For raw material $Q, 4{x} _{1}+7{x} _{2}+3{x} _{3}\le6,000$
Product $B$ requires $\frac{1}{2}$ and product $C$ requires ${\left(\frac{1}{3}\right)}^{rd}$ the time required for product $A.$
Then $\frac{t}{2}$ and $\frac{t}{3}$ are the times in hours to produce $B$ and $C$ and since $1,600$ units of $A$ will need time $1,600t$ hours, we get the constraint,
$t{x} _{1}+\frac{t}{2}{x} _{2}+\frac{t}{3}{x} _{3}\le 1,600t$ or 
${x} _{1}+\frac{{x} _{2}}{2}+\frac{{x} _{3}}{3}\le1,600$ or
$6{x} _{1}+3{x} _{2}+2{x} _{3}\le9,600$
Market demand requires
${x} _{1}\ge300, {x} _{2}\ge250,$ and ${x} _{3}\ge200$ 
Finally, since products $A,B$ and $C$ are to be produced in the ratio $3:4:5,$
${x} _{1}:{x} _{2}:{x} _{3}::3:4:5$
or $\frac{{x} _{1}}{3}=\frac{{x} _{2}}{4},$
and $\frac{{x} _{2}}{4}=\frac{{x} _{3}}{5}.$
Thus, there are two additional constraints
$4{x} _{1}-3{x} _{2}=0$ and $5{x} _{2}-4{x} _{3}=0$ where ${x} _{1},{x} _{2},{x} _{3}\ge0$ 

Find the output of the program given below if$ x = 48$
and $y = 60$
10  $ READ x, y$
20  $Let x = x/3$
30  $ Let y = x + y + 8$
40  $ z = \dfrac y4$
50  $PRINT z$
60  $End$

business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. $21$

  2. $22$

  3. $23$

  4. $24$

Show answer
Correct Option: A
Explanation:

After the step $ 10 $ READ $ x, y $, the value of $ x = 48, y = 60 $

After the step $ 20 $ Let $ x = \frac {x}{3} $, the value of $ x = \frac {48}{3} = 16 , y = 60 $

After the step $ 30 $ Let $ y = x +y + 8 $, the value of $ x = 16, y = 16 + 60 + 8 = 84   $

After the step $ 40 $ Let $ z = \frac {y}{4} $, the value of $ x = \frac {84}{4} = 21   $

Hence $ z = 21 $ is the final output printed.

Conclude from the following:
$n^2 > 10$, and n is a positive integer.
A: $n^3$
B: $50$
business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. The quantity A is may be greater or smaller than B.

  2. The quantity B is greater than A.

  3. The two quantities are equal.

  4. The relationship cannot be determined from the information given.

Show answer
Correct Option: A
Explanation:

given, $n^2 > 10$ and $n >0 $ 

multiplying both equations we get
$n^3>0$
so, it may be greater than or less than 50
Hence, quantity A is may be greater or smaller than B

For any positive real number $a$ and for any $n \in N$, the greatest value of 
$\dfrac {a^n}{1+a+a^2....a^{2n}}$ is

business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. $\dfrac 1{2n}$

  2. $\dfrac 1{2n+1}$

  3. $\dfrac 1{2n-1}$

  4. None of the above.

Show answer
Correct Option: B
Explanation:

We know that $A.M.\geq G.M.$


Therefore, $\dfrac{1+a+a^2+...+a^{2n}}{2n+1}\geq \sqrt[(2n+1)]{1*a*a^2*...*a^{2n}}$

$\implies \dfrac{1+a+a^2+....+a^{2n}}{2n+1}\geq \sqrt[(2n+1)]{a^{(1+2+...+2n)}}$

We know that sum of first $n$ numbers is $1+2+...+n=\dfrac{n(n+1)}{2}$

Therefore $1+2+...+2n=\dfrac{2n(2n+1)}{2}=n(2n+1)$

$\implies \dfrac{1+a+...+a^{2n}}{2n+1}\geq (a^{n(2n+1)})^{\dfrac{1}{2n+1}}$

$\implies \dfrac{1+a+...+a^{2n}}{2n+1}\geq a^n$

$\implies \dfrac{a^n}{1+a+...+a^{2n}}\leq \dfrac{1}{2n+1}$

Therefore the greatest value of $\dfrac{a^n}{1+a+...+a^{2n}}$ is $\dfrac{1}{2n+1}$

The rod of fixed length $k$ slides along the coordinate axes. If it meets the axes at $A(a,0)$ and $B(0,b)$, then the minimum value of $\left(a+\dfrac 1a\right) ^2+\left(b+\dfrac 1b\right) ^2$ is

business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. $0$

  2. $8$

  3. $k^2-4+\dfrac 4{k^2}$

  4. $k^2+4+\dfrac 4{k^2}$

Show answer
Correct Option: B
Explanation:

since, $A.M. \geq G.M. \implies \dfrac{a+b}{2}\geq \sqrt{ab}$


let $a=1 $ and $b=a^2$

Therefore, $\dfrac{1+a^2}{2}\geq \sqrt{a^2}$

$\implies \dfrac{1+a^2}{2}\geq \sqrt{a^2}$

$\implies \dfrac{1+a^2}{2}\geq a$

$\implies \dfrac{1+a^2}{a}\geq 2$

$\implies \dfrac{1}{a}+\dfrac{a^2}{a}\geq 2$

$\implies a+\dfrac{1}{a}\geq 2$
 squaring on both sides
$\implies (a+\dfrac{1}{a})^2\geq 4$ ------- (1)

similarly, $ (b+\dfrac{1}{b})^2\geq 4$ --------(2)

adding (1) and (2) we get

$(a+\dfrac{1}{a})^2+ (b+\dfrac{1}{b})^2\geq 4+4$

$(a+\dfrac{1}{a})^2+ (b+\dfrac{1}{b})^2\geq 8$

Therefore the minimum value is $8$

If $a>0$, then least value of $(a^3+a^2+a+1) ^2$ is

business maths linear programming problems structure of linear programming model linear programming problem operations research

  1. $64a^2$

  2. $16a^4$

  3. $16a^3$

  4. None of the above.

Show answer
Correct Option: C
Explanation:

we know that $A.M.\geq G.M.$


therefore, $\dfrac{a^3+a^2+a+1}{4}\geq \sqrt[4]{a^3*a^2*a*1}$

$\implies \dfrac{a^3+a^2+a+1}{4}\geq \sqrt[4]{a^6}$

squaring on both sides 

$\implies (\dfrac{a^3+a^2+a+1}{4})^2\geq ({a^{\dfrac{6}{4}}})^2$

$\implies ({a^3+a^2+a+1})^2\geq 16a^3$