Product pricing: solved numerical questions

1- The demand function equation is given by;

`P=5000-4Q`

Find out:

a) Marginal revenue equation

b) At what price and quantity MR will be zero?

c) At what price and quantity TR will be maximized?

d) Price elasticity of demand when TR is maximum.

Solutions:

a) To calculate MR equation let us first find out TR function.

`TR=P×Q.....(i)`

Putting the value of P in the above equation we get,

`TR=(5000-4Q)Q`

    `=5000Q-4Q^2`

Let us differentiate TR with respect to Q to find out MR.

`MR=\frac{d(TR)}{dQ}`

    `=\frac{d(5000Q-4Q^2)}{dQ}`

    `=1×5000Q^(1-1)-2×4Q^(2-1)`         (multiply coffe. value by power and subtract 1 from power)

    `=5000-8Q`

b) At what price and quantity MR will be zero?, to calculate it, let us set MR equals zero.
`MR=0`
`Or, 5000-8Q=0`
`Or, 5000=8Q`
`Or, \frac{5000}{8}=Q`
`\therefore\ Q=625`

To calculate price let us substitute Q in demand function,

`P=5000-4Q`

    `=5000-4×625`

    `=5000-2500`

    `=2500`

c) At what price and quantity TR will be maximized?, to calculate it let us set the first derivative of TR function to zero,

`\frac{d(TR)}{dQ}=0`
`Or, \frac{d(5000Q-4Q^2)}{dQ}=0`
`Or, 5000-8Q=0`
`Or, 5000=8Q`
`Or, Q=\frac{5000}{8}`
`\therefore\Q=625`

Substituting the value of Q in demand function we will get,
`P=2500`

d) Calculation of price elasticity of demand when TR is maximum.

The TR is maximum at `Q=625` and `P=2500`

`e_P=\frac{dQ}{dP}×\frac{P}{Q}`

Given the demand function `P=5000-4Q`

`4Q=5000-P`

`Q=1250-frac{1}{4}P`

`\frac{dQ}{dP}=\frac{1250-(1/4)P}{dP}`

    `=0-\frac-{1}{4}`

2-A firm has the cost and price functions given as;

`C=25+3Q^2`

`P=50-2Q`

Find:

a) Profit maximizing output,

b) Total Profit,

c) MC and MR

Solution

a) Profit is maximized at which the gap between `TR` and `TC` is maximum. Hence it is expressed as,

`π=TR-TC.............(i)`

`TC` or `C` is given as `25+3Q^2`

`TR=P✕Q...........(ii)`

Substituting `50-2Q`in equation `(ii)` for `P` we get,

`TR=(50-2Q)Q`

`TR=50Q-2Q^2`

Substituting `50-2Q^2` and `25+2Q^2` for `TR` and `TC` in equation `(i)`  , we get,

`π=50Q-2Q^2-(25+3Q^2)`

    `=50Q-2Q^2-25-3Q^2`

    `=-5Q^2+50Q-25`

Now, let's differentiate profit function with respect to `Q` and set it with `0`,

`\frac{dπ}{dQ}=0`

`\frac{d(-5Q^2+50Q-25)}{dQ}=0`

`\frac{d(-5Q^2)}{dQ}+\frac{d(50Q)}{dQ}-\frac{d(25)}{dQ}=0`

`-10Q+50-0=0`

`-10Q=-50`

`Q=\frac{-50}{-10}`

`Q=5`

Hence, profit maximizing output is 5 units.

b) Total Profit

To calculate total profit, lets plot the value of `Q` in profit function,

`π=-5Q^2+50Q-25`

    `=-5(5)^2+50(5)-25`

    `=-125+250-25`

    `=Rs.100`

c) Finding `MC` and `MR`

To find out `MC` and `MR` let's differentiate `TC` and `TR` functions with respect to `Q` and plot the value of Q,

`MC=frac\{d(TC)}{dQ}`

    `=\frac{d(25+3Q^2)}{dQ}`

    `=6Q`

Plotting the value of `Q` in `MC` function, we get,

`MC=6Q`

    `=6✕5`

    `=Rs.30`

`MR=\frac{d(TR)}{dQ}`

    `=\frac{d(50Q-2Q^2)}{dQ}`

    `=50-4Q`

Plotting the value of `Q` in `MR` function, we get,

`MR=50-4Q`

    `=50-4✕5`

    `=50-20`

    `=Rs.30`

3- A firm has the following demand and cost functions;

`TC=Q^3-8Q^2+57Q+2`

`0.5Q=45-P`

Now, Find out:

a) Profit maximizing output,

b) Selling price,

c) Total profit

Solution:

a) Given the cost and demand function;

`TC=Q^3-8Q^2+57Q+2`

`0.5Q=45-P`

`P=45-0.5Q`

`TR=P✕Q`

    `=(45-0.5Q)Q`

    `=45Q-0.5Q^2`

Equilibrium conditions are;

`MR=MC`

`MC>0`

To find out `MR` let's take first order derivative of `TR` function.

`MR=\frac{d(TR)}{dQ}`

    `=\frac{d(45Q-0.5Q^2)}{dQ}`

    `=45-Q`

To find out `MC` let's take first order derivative of `TC` function.

`MC=\frac{d(TC)}{dQ}`

    `=\frac{d(Q^3-8Q^2+57Q+2)}{dQ}`

    `=3Q^2-16Q+57`

Equating `MR` and `MC` functions;

`45-Q=3Q^2-16Q+57`

`\Or,45-57-Q+16Q-3Q^2=0`

`Or,-3Q^2+15Q-12=0`

Dividing both sides by `(-3)` ;

`Q^2-5Q+4=0`

Factorizing the equation;

`Q^2-Q-4Q+4=0`

`Q(Q-1)-4(Q-1)=0`

`(Q-1)(Q-4)=0`

`Q-1=0`    `Q-4=0`

`Q=1`        `Q=4`

`Q= 1,\or\4`  it's undermined,

To determine the profit maximizing output, let's take the second order derivate of `MC` function;

`MC=3Q^2-16Q+57`

  

`MC'=\frac{d(MC)}{dQ}`

    `=\frac{d(3Q^2-16Q+57)}{dQ}`

    `6Q-16`

If `Q=1` then,

`MC=6✕1-16`

    `=-10`

    `-10<0` (negative)

If `Q=4` then,

`MC=6✕4-16`

    `24-16`

    `8` 

    `8>0` (positive)

Hence profit maximizing output;

`Q=4`

b) Selling price

To calculate setting price, let's put the value of `Q` in price function,

`P=45-0.5Q`

    `=45-0.5✕4`

    `=45-2`

    `=Rs, 43` per unit

c) Total profit

`π=TR-TC`

    `=45Q-0.4Q^2-(Q^3-8Q^2+57Q+2)`

    `=45(4)-0.5(4)^2-(4)^3+8(4)^2-57(4)-2`

    `=180-8-64+128-228-2`

    `=308-302`

    `=Rs. 6`

Hence;

Profit maximizing output `Q=4` units,

Selling price `P=Rs. 43`

Total profit `π=Rs. 6`

4-A monopolist faces the following demand and cost functions.

`P=100-Q`

`C=25+10Q+2Q^2`

Now, Find:

a) Price of the product,

b) Profit maximizing output,

c) Total profit

Solution:

a) Before we calculate the price let us first find out the profit maximizing output.

b) We know that profit is maximized at which the gap between `TR` and `TC` is maximum. Hence profit function is expressed as given below;

Profit `π=TR-TC`

Here,

`TR=P×Q`

    `=(100-Q)×Q`

    `=100Q-Q^2`

`TC=C=25+10Q+2Q^2`

Putting `TR` and `TC` function in profit function,

`π=100Q-Q^2-(25+10Q+2Q^2)`

    `=100Q-Q^2-25-10Q-2Q^2`

    `=-3Q^2+90Q-25`

For profit maximization,  `\frac{dπ}{dQ}=0`  and `\frac{d^2π}{dQ^2}<0`

 Taking first order derivative of profit function we get,

`\frac{dπ}{dQ}=0`

`Or, \frac{d(-3Q^2+90Q-25)}{dQ}=0`

`Or, -6Q+90=0`

`Or, -6Q=-90`

`Or, Q=90/6`

`\therefore` `Q=15`

Taking second order derivative of profit function,

`\frac{d^2π}{dQ^2}=-6<0`

Hence, `Q=15` units.

Putting the value of `Q` in demand function we get,

`P=100-Q`

    `=100-15`

    `=Rs, 85`

c) Total profit

`π=TR-TC`

`π=100Q-Q^2-(25+10Q+2Q^2)`

`=100Q-Q^2-25-10Q-2Q^2`

    `=-3Q^2+90Q-25`

Putting the value of `Q` in profit function we get total profit,

`π=-3(15)^2+90(15)-25`

`=-675+1350-25`

`=1350-700`

`=Rs,\650`

Hence,

Price of output `P=85`

Profit maximizing output `Q=15`

Total profit `π=650`

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