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Questions Related to business maths

Apply linear programming to this problem. A firm wants to determine how many units of each of two products (products D and E) they should produce to make the most money. The profit in the manufacture of a unit of product D is $100 and the profit in the manufacture of a unit of product E is $87. The firm is limited by its total available labor hours and total available machine hours. The total labor hours per week are 4,000. Product D takes 5 hours per unit of labor and product E takes 7 hours per unit. The total machine hours are 5,000 per week. Product D takes 9 hours per unit of machine time and product E takes 3 hours per unit. Which of the following is one of the constraints for this linear program?

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

  1. $5 D + 7 E≤ 5,000$

  2. $9 D + 3 E ≥4,000$

  3. $5 D + 7 E = 4,000$

  4. $5 D + 9 E ≤5,000$

  5. $9 D + 3 E ≤5,000$

Show answer
Correct Option: E
Explanation:

Given, product D takes 5 hours per unit of labour, and
product E takes 7 hours per unit of labour.
Therefore, to produce D units of product D takes $5D$ hours and
to produce E units of product E takes $7E$ hours 
Given, total labour hours per week are $4000$ hours.
Hence, $5D+7E\leq 4000$

Given, product D takes 9 hours per unit of machine time, and
product E takes 3 hours per unit of machine time.
Therefore, to produce D units of product D takes $9D$ hours and
to produce E units of product E takes $3E$ hours 
Given, total machine hours per week are $5000$ hours.
Hence, $9D+3E\leq 5000$

To write the dual; it should be ensured that  
I. All the primal variables are non-negative.
II. All the bi values are non-negative.
III. All the constraints are $≤$ type if it is maximization problem and $≥$ type if it is a minimization problem.

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  1. I and II

  2. II and III

  3. I and III

  4. I, II and III

Show answer
Correct Option: C
Explanation:

To write the dual, then all the primal variables must be non-negative.

All the constraints are $\leq$ type if it ia maximization problem and $\geq$ type if it is a minimization problem.

If $x=\log _{2^2}2+\log _{2^3}2^2+\log _{2^4}2^3......+\log _{2^{n+1}}2^n+$, then the minimum value of $x$ will be-

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

  1. $\left(\dfrac 1{n+1}\right)^{\tfrac 1n}$

  2. $n\left(\dfrac 1{n+1}\right)^{\tfrac 1n}$

  3. $\left(\dfrac n{n+1}\right)^{\tfrac 1n}$

  4. None of the above.

Show answer
Correct Option: B
Explanation:
given $x={log _{2^2}2+log _{2^3}{2^2}+log _{2^4}{2^3}+...+log _{2^{n+1}}2^n}$

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

$\implies \dfrac{log _{2^2}2+log _{2^3}{2^2}+log _{2^4}{2^3}+...+log _{2^{n+1}}2^n}{n}\geq \sqrt[n]{log _{2^2}2*log _{2^3}{2^2}*log _{2^4}{2^3}*...*log _{2^{n+1}}{2^n}}$

$\implies \dfrac{x}{n}\geq \sqrt[n]{\dfrac{log2}{log2^2}*\dfrac{log2^2}{log2^3}*\dfrac{log2^3}{log2^4}*...*\dfrac{log2^n}{log2^{n+1}}}$

$\implies x\geq n\sqrt[n]{\dfrac{log2}{log2^{n+1}}}$

$\implies x\geq n({\dfrac{log2}{(n+1)log2}})^{\dfrac 1n}$

$\implies x\geq n({\dfrac{1}{n+1}})^{\dfrac 1n}$

therefore the minimum value of $x$ is $ n({\dfrac{1}{n+1}})^{\dfrac 1n}$

If $a,b >0$, $a+b=1$, then the least value of $(1+\dfrac 1a)(1+\dfrac 1b)$, is

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

  1. $3$

  2. $6$

  3. $9$

  4. $12$

Show answer
Correct Option: C
Explanation:
Given, $a+b=1$
we know that, $A.M.\geq G.M.$

$\implies \dfrac{a+b}{2}\geq \sqrt{ab}$

$\implies \dfrac{1}{2}\geq \sqrt{ab}$

$\implies \sqrt{ab}\leq \dfrac{1}{2}$

squaring on both sides

$\implies ab \leq \dfrac{1}{4}$  --------------(1)

Similarly

$\implies \dfrac{1+a+1+b}{2}\geq \sqrt{(1+a)(1+b)}$

$\implies \dfrac{2+(a+b)}{2}\geq \sqrt{(1+a)(1+b)}$

$\implies \dfrac{2+1}{2}\geq \sqrt{(1+a)(1+b)}$

$\implies \dfrac{3}{2}\geq \sqrt{(1+a)(1+b)}$

squaring on both sides

$\implies \dfrac{1}{(1+a)(1+b)}\leq \dfrac{4}{9}$  ---------------(2)

multiplying (1) and (2) we get

$\implies \dfrac{ab}{(1+a)(1+b)}\leq \dfrac{1}{4}*\dfrac{4}{9}$

$\implies \dfrac{ab}{(1+a)(1+b)}\leq \dfrac{1}{9}$

$\implies \dfrac{1}{(1+a)(1+b)}\leq \dfrac{1}{9ab}$

$\implies \dfrac{(1+a)(1+b)}{ab}\geq 9$

$\implies \dfrac{(1+a)}{a}*\dfrac{(1+b)}{b}\geq 9$

$\implies (1+\dfrac{1}{a})(1+\dfrac{1}{b})\geq 9$

Therefore, the minimum value of $ (1+\dfrac{1}{a})(1+\dfrac{1}{b})$ is $ 9$

If $l,m,n$ be three positive roots of the equation $x^3-ax^2+bx+48=0$, then the minimum value of $\dfrac 1l +\dfrac 2m+\dfrac 3n$ is

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

  1. $1$

  2. $2$

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

  4. $\dfrac 52$

Show answer
Correct Option: C
Explanation:

we Know that, $A.M.\geq G.M.$

$\implies \dfrac{a+b+c}{3}\geq \sqrt[3]{abc}$

let $a=\dfrac{1}{l}, b=\dfrac{2}{m}, c=\dfrac{3}{n}$

Therefore,

$\dfrac{1}{3}(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq \sqrt[3]{(\dfrac{1\times2\times3}{lmn})}$


$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{1\times2\times3}{lmn})}$

Given, the roots of the polynomial $x^3-ax^2+bx+48=0$ are $l,m,n$
Therefore, the product of the roots $lmn=-(\dfrac{48}{1})=-48$

Substituting $lmn=-48$ in the above equation

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{6}{-48})}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{1}{-8})}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(-\dfrac{1}{2})^3}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times(-\dfrac{1}{2})$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq (-\dfrac{3}{2})$

therefore, the minimum value is $-\dfrac{3}{2}$

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