Tag: minimization of cost function and maximization of revenue function and profit function

Questions Related to minimization of cost function and maximization of revenue function and profit function

A firm $ABC$ starts producing pens and finds that the production cost of each pen is Rs $10$, and the fixed expenditures of production is Rs. $4500$. If each pen is sold for Rs. $25$, find break-even point.
business mathematics and statistics applications of calculus revenue functions from marginal revenue functions minimization of cost function and maximization of revenue function and profit function integral calculus – ii

  1. $x=200$

  2. $x=100$

  3. $x=400$

  4. $x=300$

Show answer
Correct Option: D
Explanation:
Manufacturing cost $=10$
Number of pens $=10$
Total manufacturing cost $=10x+4500$
Selling cost $=25x$
Break even value $\Rightarrow $ manufacturing cost $=$ Selling cost 
$\Rightarrow 10x+4500=25x$
$x=300$
A company sells its product at the rate of Rs. $6$ per unit. The variable costs are estimated to run $25\%$ of the total revenue received. If the fixed costs for the product are Rs. $4500$. Find the total cost function.
business mathematics and statistics applications of calculus revenue functions from marginal revenue functions minimization of cost function and maximization of revenue function and profit function integral calculus – ii

  1. $4500x$

  2. $4500x+\dfrac{1}{2}x$

  3. $4500x+\dfrac{3}{2}x$

  4. $4500x+\dfrac{5}{2}x$

Show answer
Correct Option: C
Explanation:

$\Rightarrow$  Here, price per unit $(p)=Rs.6$ and fix cost is $Rs.4500$.

$\Rightarrow$  $Total\,\, revenue\,\,R(x)=p.x=6x$  where $x$ is number of unit sold.
$\Rightarrow$  $Total\, \,cost\,\, function\,\,C(x)=4500x+\dfrac{25}{100}R(x)$
$\Rightarrow$  $C(x)=4500x+\dfrac{25}{100}\times 6x$

$\Rightarrow$  $C(x)=4500x+\dfrac{3}{2}x$ 

A company sells its product at the rate of Rs. $6$ per unit. The variable costs are estimated to run $25\%$ of the total revenue received. If the fixed costs for the product are Rs. $4500$. Find the profit function.
business mathematics and statistics applications of calculus revenue functions from marginal revenue functions minimization of cost function and maximization of revenue function and profit function integral calculus – ii

  1. $6x-\dfrac{3}{2}x-4500$

  2. $6x-\dfrac{1}{2}x-4500$

  3. $6x+\dfrac{3}{2}x-4500$

  4. $6x-\dfrac{5}{2}x+4500$

Show answer
Correct Option: A
Explanation:

$\Rightarrow$  Here, price per unit $(p)=Rs.6$

$\Rightarrow$  Total revenue $R(x)=p.x=6x$  where $x$ is the number of unit sold.
$\Rightarrow$  Cost function $C(x)=4500+\dfrac{25}{100}R(x)$

$\Rightarrow$  Cost function $C(x)=4500+\dfrac{25}{100}\times 6x$

$\Rightarrow$   $C(x)=4500+\dfrac{3}{2}x$

$\Rightarrow$   Profit function $P(x)=R(x)-C(x)$

$\Rightarrow$  $P(x)=6x-(4500+\dfrac{3}{2}x)$

$\therefore$    $P(x)=6x-\dfrac{3}{2}x-4500$

Find the elasticity of supply when price $5$ units. Supply function is given by $q = 25 - 4p +p^2$

business mathematics and statistics applications of calculus revenue functions from marginal revenue functions minimization of cost function and maximization of revenue function and profit function integral calculus – ii

  1. $1$

  2. $2$

  3. $0$

  4. Cannot be determined

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Correct Option: A
Explanation:
$q=25-4p+p^{2}$
Elasticity of supply $=\cfrac{\cfrac{dq}{q}}{\cfrac{dp}{p}}=\cfrac{(-4+2p)\times p}{(25-4p+p^{2})}$
When $p=5$ units
Elasticity of supply $=\cfrac{(-4+10)\times 5}{(25-20+25)}=1$

Find the elasticity of supply for supply function $x = 2p^2+5$, when $p=3$.

business mathematics and statistics applications of calculus revenue functions from marginal revenue functions minimization of cost function and maximization of revenue function and profit function integral calculus – ii

  1. $\dfrac{23}{36}$

  2. $\dfrac{36}{23}$

  3. $\dfrac{63}{32}$

  4. None of these

Show answer
Correct Option: B
Explanation:
Elasticity of supply$\Rightarrow\cfrac{\cfrac{+dx}{x}}{\cfrac{dp}{p}}$$\Rightarrow\cfrac{(4p)p}{2p^{2}+5}$
Elasticity of supply when $p=3\Rightarrow\cfrac{36}{23}$

A village road is to be constructed by a team of $250$ workers. After $12$ days it was found that only $2/7^{th}$ part of the work was complete. To complete the rest in another $25$ days, how many more workers should be employed?

business mathematics and statistics applications of calculus revenue functions from marginal revenue functions minimization of cost function and maximization of revenue function and profit function integral calculus – ii

  1. $53$

  2. $52$

  3. $55$

  4. $50$

Show answer
Correct Option: D
Explanation:

Let the total unit of work be $7$ units.
According to question,
$\dfrac {M _{1}D _{1}}{W _{1}} = \dfrac {M _{2}D _{2}}{W _{2}} \Rightarrow \dfrac {250\times 12}{2} = \dfrac {M _{2}\times 25}{(7 - 2)} \Rightarrow M _{2} = 300$
Number of extra workers needed $= 300 - 250 = 50$.

A railway half ticket costs half the full fare but reservation charge is same. One reserved ticket from Ranchi to Howrah cost $Rs.1720$ and one full and one half ticket (both reserved) costs $Rs.2610$. Thus, the reservation charge is

business mathematics and statistics applications of calculus revenue functions from marginal revenue functions minimization of cost function and maximization of revenue function and profit function integral calculus – ii

  1. $Rs.60$

  2. $Rs.30$

  3. $Rs.40$

  4. None

Show answer
Correct Option: A
Explanation:

Let the full ticket cost $ Rs  \ x $and reservation charge be $ y $
As per given statements,
$ x + y = 1720 $    -(1)
And $ x + y + \frac {x}{2} + y = 2610 => \frac {3x}{2} + 2y = 2610 $ --- (2)

Multiplying equation  $ (1) $ with $ 2 $ we get, $ 2x + 2y
= 3440 $ ----- equation $ (3) $

Subtracting equation $ (2) $

from $ (3) $, we get $ \frac{x}{2} = 830 => x = 1660 $

Substituting $ x = 1660 $ in the equation $ (1) $, we get $ 1660 +y = 1720 => y = 60 $ 

Thus reservation charges were Rs $60 $