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Solved numerical questions on short run cost


1-Suppose a firm has a cost function given by `TC=10+5Q+Q^2`
Find:
a) Total fixed cost of the firm,
b) Average variable cost and marginal cost at `Q=10`,

Solution,
a) Total fixed cost (TFC) equals total cost (TC) at zero output level, and it is expressed as,
`TFC=TC`, when `Q=0`
Hence,
`TFC=TC=10+5Q+Q^2`
    `=10+5(0)+(0)^2`
    `=10+0+0`
    `=10`
b) Average variable cost (AVC);
To calculate AVC, let us first calculate total variable cost (TVC),
`TVC=TC-TFC`
    `=10+5Q+Q^2-10`
    `=5Q+Q^2`

`AVC=\frac{TVC}{Q}`
Substituting `5Q+Q^2` for `TVC` we get,
`AVC=frac\{5Q+Q^2}{Q}`
    `=5+Q`
    `=5+10`  `(\Given, Q=10)`
    `=15`

To calculate `MC` let us differentiate `TC` function with respect to `Q`,
`MC=\frac{d(TC)}{dQ}`
    `=\frac{d(10+5Q+Q^2)}{dQ}`
    `=0+5+2×Q^(2-1)`
    `=5+2Q`
    `=5+2(10)`   `(Given, Q=10)`
    `=25`
Hence,
`TFC=10`
`AVC=15`
`MC=25`

2- The short run total cost is `TC=300+4Q-0.4Q^2+0.2Q^3`. If `Q=10`, Find; TFC,  AVC,  AFC and  MC.

Solution

Calculation of TFC,
TFC equals TC at zero output level and is expressed as,
`TFC=TC`  when `Q=0`
`TFC=TC=300+4Q-0.4Q^2+0.2Q^3`
    `=300+4(0)-0.4(0)^2+0.2(0)^3`
    `=300+0-0+0`
    `=300`
Calculation of AVC at `Q=10`,
To calculate AVC at `Q=10` let us first calculate TVC,
`TVC=TC-TFC`
    `=300+4Q-0.4Q^2+0.2Q^3-300`  `(\because\, TFC=300)`
    `=4Q-0.4Q^2+0.2Q^3`
`AVC=\frac{TVC}{Q}`
    `=\frac{4Q-0.4Q^2+0.2Q^3}{Q}`
    `=\frac{Q(4-0.4Q+0.2Q^2)}{Q}`
    `=4-0.4Q+0.2Q^2`
    `=4-0.4(10)+0.2(10^2)`       `(\Given, Q=10)`
    `=4-4+20`
    `=20`

Calculation of AFC at `Q=10`,
`AFC=\frac{TFC}{Q}`
    `=\frac{300}{10}`         `(\Given, Q=10)`
    `=30`

Calculation of MC at `Q=10`
To calculate MC, let us differentiate TC function with respect to Q,
`MC=\frac{d(TC)}{dQ}`
    `=\frac{d(300+4Q-0.4Q^2+0.2Q^3)}{dQ}`
    `=0+1×4Q^(1-1)-2×0.4Q^(2-1)+3×0.2Q^(3-1)`
    `=4-0.8Q+0.6Q^2`
    `=4-0.8(10)+0.6(10)^2`
    `=4-8+60`
    `=56`
Hence,
`TFC=300`
`AVC=20`
`AFC=30`
`MC=56`

3- A company has the following variable cost function.
`TVC=250Q-8Q^2+0.5Q^3`
If the company's fixed costs are 200 lakh, find out:
a) Total cost function,
b) Marginal cost function,
c) Average variable cost function,
d) Average total cost function,
e) At what output levels AVC and MC will be minimum,

Solutions:

a) Calculation of TC function,

TC is the sum of TFC and TVC, and it is expressed as,
`TC=TFC+TVC`
    `=200+250Q-8Q^2+0.5Q^3`         (Given the value of  `TFC` and `TVC`)

b) Calculation of MC function,

To calculate `MC` let us differentiate `TC` function with respect to `Q` . 
`MC=\frac{d(TC)}{dQ}`
    `=frac\{d(200+250Q-8Q^2+0.5Q^3)}{dQ}`
    `=0+250-16Q+1.5Q^2`
    `=250-16Q+1.5Q^2`

c) Calculation of AVC function,

`AVC=\frac{TVC}{Q}`
    `=\frac{250Q-8Q^2+0.5Q^3}{Q}`
    `=\frac{Q(250-8Q+0.5Q^2)}{Q}`
    `=250-8Q+0.5Q^2`

d) Calculation of AC,

`AC=\frac{TC}{Q}`
`=\frac{200+250Q-8Q^2+0.5Q^3}{Q}`
`=\frac{200}{Q}+250-8Q+0.5Q^2`

e) Calculation of output levels at which AVC and MC are minimum,

To calculate the output level at which AVC is minimum, take the first derivative of AVC function and set the derivative equal to zero.
We have `AVC=250-8Q+0.5Q^2`
`\frac{d(AVC)}{dQ}=0`
`Or, \frac{d(250-8Q+0.5Q^2)}{dQ}=0`
`Or, 0-8+Q=0`
`Or, Q=8`

To calculate the output level at which MC is minimum, take the first derivative of MC function and set the derivative equal to zero.
We have `MC=250-16Q+1.5Q^2`
`\frac{d(MC)}{dQ}=0`
`Or, \frac{d(250-16Q+1.5Q^2)}{dQ}=0`
`Or, 0-16+3Q=0`
`Or, 3Q=16`
`Or, Q=\frac{16}{3}`
        `=5.33`

4- A firm producing cricket bats has a production function given by,
`Q=2√KL`
Its short run fixed cost `K=100,` rental rate for `K` is `$1` and wage rate is `$4.`
Find:
a) TC and AC
b) TC, AC and MC for producing 20 cricket bats.

Solutions:
Given the production function as;
`Q=2√KL`
Putting value of `K` in production function we get,
`Q=2√100L`
    `=2×10√L`
    `=20√L`
Taking square on both sides,
`Q^2=(20√L)^2`
`Q^2=400L`
`L=\frac{Q^2}{400}`

Cost function `C=wL+rK`
Putting the values of w, L, r and K in cost function,
`C=4(\frac{Q^2}{400})+1×100`
    `=\frac{Q^2}{100}+100`
    `\therefore\TC=\frac{Q^2}{100}+100`

Calculating AC
`AC=\frac{TC}{Q}`
`=\frac{Q^2/100+100}{Q}`
`=\frac{Q}{100}+\frac{100}{Q}`

Calculation of TC, AC and MC for producing 20 cricket bats.
Since, `TC=\frac{Q^2}{100}+100`
    `=\frac{20^2}{100}+100`        (producing 20 bats)
    `=\frac{400}{100}+100`
    `=4+100`
    `=$104`

Since `AC=\frac{Q}{100}+\frac{100}{Q}`
    `=\frac{20}{100}+\frac{100}{20}`            (producing 20 bats)
    `=frac{20+500}{100}`
    `=frac\{520}{100}`
    `=$5.2`

`MC=\frac{d(TC)}{dQ}`
    `=\frac{d(Q^2/100+100)}{dQ}`
    `=\frac{2Q}{100}+0`
    `=\frac{Q}{50}`
    `=\frac{20}{50}`             (producing 20 bats)
    `=$0.4`

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