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Revenue functions from marginal revenue functions - class-XII

Description: revenue functions from marginal revenue functions
Number of Questions: 17
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Tags: applications of calculus differentiation business mathematics and statistics statistics integral calculus – ii maths
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The demand function if $p = 60 + 2D - 10D^2$, the rate of charge in price with respect to demand is ______

  1. $\dfrac{60}{D} + 2 - 10D$

  2. $2 - 20D$

  3. $2 - 40D$

  4. None of these


Correct Option: B

Given the total cost function, $TC = a+bQ+cQ^2+dQ^3$, Find the Marginal cost.

  1. $b+2cQ+3dQ^2$

  2. $a + b+2cQ+3dQ^2$

  3. $a + 2b+2cQ+3dQ^2$

  4. None


Correct Option: A
Explanation:

$\Rightarrow$  We have $TC=a+bQ+cQ^2+dQ^3$

$\Rightarrow$  $Marginal\,cost = \dfrac{d}{dQ}TC$

$\Rightarrow$  $Marginal\, cost=\dfrac{d}{dQ}(a+bQ+cQ^2+dQ^3)$

$\therefore$    $Marginal\,cost=b+2cQ+3dQ^2$

 The total cost function is $TC = 12x + 2x^2$. Find the $MC$.

  1. $12-4x$

  2. $4x - 12$

  3. $12 + 4x$

  4. None of these


Correct Option: C
Explanation:

$MC = \dfrac{\mathrm{d} TC}{\mathrm{d} x}$

$TC = 12x + 2x^{2}$
$\Rightarrow \dfrac{\mathrm{d} TC}{\mathrm{d} x}=12+4x$
$\Rightarrow MC = 12 + 4x$

$y = 48x - 2x^2$
where, $y=$ Total revenue $ $ $.
$x = $ Output
At what output is total revenue a maximum?

  1. $2$

  2. $12$

  3. $48$

  4. $4$


Correct Option: B
Explanation:

$\Rightarrow$  We have given, $y=48x-2x^2$

$\Rightarrow$  Differentiate on both sides w.r.t $x$ we get,
$\Rightarrow$  $\dfrac{dy}{dx}=48-4x$          --- ( 1 )
$\Rightarrow$  Let $\dfrac{dy}{dx}=0$
$\Rightarrow$  $48-4x=0$
$\Rightarrow$  $4x=48$
$\Rightarrow$  $x=12$ is a turning point.
$\Rightarrow$  $\dfrac{d^2y}{dx^2}=-4$         [Differentiate ( 1 ) on both sides]
$\Rightarrow$  So, the turning point is a maximum 

The cost, in dollars, of producing $x$ gallons of detergent is given by
$C(x)=350+20x0.08x^2 + 0.0004x^3$
What is a formula for the marginal cost function $C'(x)$

  1. $C(x)=200.16x^2+0.0012x^2$

  2. $C(x)=200.16x+0.0012x^2$

  3. $C(x)=20x0.16x+0.0012x^2$

  4. $C(x)=20x0.16x^2+0.0012x^2$


Correct Option: B
Explanation:
$c'(x)=\cfrac{d}{dx}(c(x))$
$=2\times(100+0.8)x+3\times0.0004x^{2}$
$=200.16x+0.0012x^{2}$

If we differentiate the cost function: $y = \dfrac{x^4}4 + 2x^2$, we get

  1. $\dfrac{x^3}4 + 4x$

  2. $4x^3+4x$

  3. $x^3 + 4x$

  4. $4x^3+2x^2$


Correct Option: C
Explanation:

Suppose: $y = ax^n$
Then, $dy = nax^{n-1}$
Remember
$\dfrac{x^4}{4} = \dfrac{1}{4}x^4$
So differentiating both terms in the expression for using the formula gives:
$\dfrac{dy}{dx} = x^3+4x$

A ABC firms start 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 , determine cost function.

  1. 4500+10x

  2. 10+4500x

  3. 25+10x

  4. 10+25x


Correct Option: A
Explanation:

The cost function equation is expressed as C(x)= FC + V(x), where C equals total production cost, FC is total fixed costs, V is variable cost and x is the number of units.
$4500$ =fixed expenditures
$10x$ = variable cost
C(X)= $4500+10x$

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 break even point.
  1. $1000$

  2. $2000$

  3. $3000$

  4. $3500$


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 $P(x)=R(x)-C(x)$

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

$\therefore$    $P(x)=6x-\dfrac{3}{2}x-4500$
$\Rightarrow$   At break even point $P(x)=0$
$\Rightarrow$   $6x-\dfrac{3}{2}x-4500=0$

$\Rightarrow$   $\dfrac{12x-3x}{2}-4500=0$

$\Rightarrow$   $x=\dfrac{9000}{9}=1000$
$\Rightarrow$  Hence, $x=1000$ is break even point.

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 revenue function.

  1. $6x$

  2. $4x$

  3. $4500x$

  4. $6x+4$


Correct Option: A
Explanation:
$\Rightarrow$  If x is the number of units of certain product sold at a rate of $Rs.p$ per unit, then the amount derived from the sale of $x$ units of a product is the total revenue. 
$\Rightarrow$  Here, price per unit is $Rs.6$, so $p=6$
$\Rightarrow$  $Total\,revenue\,\,R(x)=p.x$
$\Rightarrow$  $Total\,revenue\,\,R(x)=6x$
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 Revenue function
  1. $25x$

  2. $10x$

  3. $4500x$

  4. $4500+10x$


Correct Option: A
Explanation:

R(x)= ( price per unit)*(number of units produced or sold)
Revenue function = $25 \times x$ (number of units)
Revenue function =$25x$

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.
  1. $x=200$

  2. $x=100$

  3. $x=400$

  4. $x=300$


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.
  1. $4500x$

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

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

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


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

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

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

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


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 $q = 25 - 4p +p^2$

  1. $1$

  2. $2$

  3. $0$

  4. Cannot be determined


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 $p=3$.

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

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

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

  4. None of these


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?

  1. $53$

  2. $52$

  3. $55$

  4. $50$


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

  1. $Rs.60$

  2. $Rs.30$

  3. $Rs.40$

  4. None


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 $

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