# A manufacturer determines that his total cost function is$c = \dfrac{{{q^3}}}{3} + 2q + 300$. Where q is the number of units produced.

$(a)$Find the marginal cost function

$(b)$Find the average cost function

$(c)$Find the level of output at which the average cost is minimum

Last updated date: 25th Mar 2023

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Answer

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Hint- Use the definitions of marginal cost, average cost, and standard procedure to find level output at average cost which is minimum.

Now the total cost is given as$c = \dfrac{{{q^3}}}{3} + 2q + 300$.

So firstly let’s calculate for (a) Marginal cost function

Marginal cost function = $\dfrac{{dc}}{{dq}} = \dfrac{d}{{dq}}\left( {\dfrac{{{q^3}}}{3} + 2q + 300} \right)$

The derivative gives us

$\dfrac{{dc}}{{dq}} = {q^2} + 2$ As derivative of ${x^n} = n{x^{n - 1}}$

So the marginal cost function is${q^2} + 2$.

Now let’s calculate for average cost function

Average cost function = $\dfrac{{total{\text{ cost}}}}{{number{\text{ of units produced}}}}$

Average cost = $\dfrac{c}{q} = \dfrac{{\dfrac{{{q^3}}}{3} + 2q + 300}}{q} = \dfrac{{{q^2}}}{3} + 2 + \dfrac{{300}}{q}$

So average cost function is $\dfrac{{{q^2}}}{3} + 2 + \dfrac{{300}}{q}$…………………………………… (1)

Now to find the level output at which average cost is minimum we simply need to put the derivative of the function of average cost equal to 0

That is $\dfrac{{d(Average{\text{ cost)}}}}{{dq}} = 0$

So we have using equation 1

$\dfrac{d}{{dq}}\left( {\dfrac{{{q^2}}}{3} + 2 + \dfrac{{300}}{q}} \right) = 0$

The derivative of this quantity is $\dfrac{{2q}}{3} - \dfrac{{300}}{{{q^2}}} = 0$

On further solving we get

$2{q^3} - 900 = 0$

Or ${q^3} = 450$

So the value of $q{ = ^3}\sqrt {450} $

Now let’s verify that this q corresponds to the minimum of the average cost function.

So $\dfrac{{{\partial ^2}(average{\text{ cost)}}}}{{\partial {q^2}}} < 0$

Let’s substitute the values we get $\dfrac{{{\partial ^2}}}{{\partial {q^2}}}\left( {\dfrac{{{q^2}}}{3} + 2 + \dfrac{{300}}{q}} \right) > 0$

Double derivative of this quantity is $\dfrac{2}{3} + \dfrac{{600}}{{{q^3}}}$…………………………………. (2)

Now on substitution of $q{ = ^3}\sqrt {450} $in above equation 2, equation 2 becomes $\dfrac{2}{3} + \dfrac{{600}}{{{{\left( {^3\sqrt {450} } \right)}^3}}} > 0$

Clearly it is positive hence it’s verified.

Note- The problem statement of this type is purely based upon the definition conceptuality of marginal function, average cost function and minimization of average cost function. The standard procedure as mentioned above leads to the answers in such types of problems.

Now the total cost is given as$c = \dfrac{{{q^3}}}{3} + 2q + 300$.

So firstly let’s calculate for (a) Marginal cost function

Marginal cost function = $\dfrac{{dc}}{{dq}} = \dfrac{d}{{dq}}\left( {\dfrac{{{q^3}}}{3} + 2q + 300} \right)$

The derivative gives us

$\dfrac{{dc}}{{dq}} = {q^2} + 2$ As derivative of ${x^n} = n{x^{n - 1}}$

So the marginal cost function is${q^2} + 2$.

Now let’s calculate for average cost function

Average cost function = $\dfrac{{total{\text{ cost}}}}{{number{\text{ of units produced}}}}$

Average cost = $\dfrac{c}{q} = \dfrac{{\dfrac{{{q^3}}}{3} + 2q + 300}}{q} = \dfrac{{{q^2}}}{3} + 2 + \dfrac{{300}}{q}$

So average cost function is $\dfrac{{{q^2}}}{3} + 2 + \dfrac{{300}}{q}$…………………………………… (1)

Now to find the level output at which average cost is minimum we simply need to put the derivative of the function of average cost equal to 0

That is $\dfrac{{d(Average{\text{ cost)}}}}{{dq}} = 0$

So we have using equation 1

$\dfrac{d}{{dq}}\left( {\dfrac{{{q^2}}}{3} + 2 + \dfrac{{300}}{q}} \right) = 0$

The derivative of this quantity is $\dfrac{{2q}}{3} - \dfrac{{300}}{{{q^2}}} = 0$

On further solving we get

$2{q^3} - 900 = 0$

Or ${q^3} = 450$

So the value of $q{ = ^3}\sqrt {450} $

Now let’s verify that this q corresponds to the minimum of the average cost function.

So $\dfrac{{{\partial ^2}(average{\text{ cost)}}}}{{\partial {q^2}}} < 0$

Let’s substitute the values we get $\dfrac{{{\partial ^2}}}{{\partial {q^2}}}\left( {\dfrac{{{q^2}}}{3} + 2 + \dfrac{{300}}{q}} \right) > 0$

Double derivative of this quantity is $\dfrac{2}{3} + \dfrac{{600}}{{{q^3}}}$…………………………………. (2)

Now on substitution of $q{ = ^3}\sqrt {450} $in above equation 2, equation 2 becomes $\dfrac{2}{3} + \dfrac{{600}}{{{{\left( {^3\sqrt {450} } \right)}^3}}} > 0$

Clearly it is positive hence it’s verified.

Note- The problem statement of this type is purely based upon the definition conceptuality of marginal function, average cost function and minimization of average cost function. The standard procedure as mentioned above leads to the answers in such types of problems.

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