Question

The total cost $C(x)$ in rupees associated with the production of $x$ units of an item is given by$C(x)=0.007{{x}^{3}}-0.003{{x}^{2}}+15x+4000$.
Find the marginal revenue when $x=17$.

Answer 1

Hint: For finding marginal revenue or marginal cost differentiate the total cost and after differentiating substitute$x=17$ you will get the marginal cost.

Complete step-by-step answer:
Marginal cost is the change in the total cost that arises when the quantity produced is incremented by one unit; that is, it is the cost of producing one more unit of a good.

 Intuitively, marginal cost at each level of production includes the cost of any additional inputs required to produce the next unit. At each level of production and time period being considered, marginal costs include all costs that vary with the level of production, whereas other costs that do not vary with production are fixed and thus have no marginal cost.

For example, the marginal cost of producing an automobile will generally include the costs of labor and parts needed for the additional automobile but not the fixed costs of the factory that have already been incurred.

In practice, marginal analysis is segregated into short and long-run cases, so that, over the long run, all costs (including fixed costs) become marginal.

Marginal cost pricing is not a matter of merely lowering the general level of prices with the aid of a subsidy; with or without subsidy it calls for a drastic restructuring of pricing practices, with opportunities for very substantial improvements in efficiency at critical points.
If the cost function$C$is continuous and differentiable, the marginal cost$MC$ is the first derivative of the cost function with respect to the output quantity$Q$.

Complete step-by-step answer:

$MC(Q)=\dfrac{dC}{dQ}$

So in question let the marginal cost be$MC$,

And the total cost be$C(x)$where$x$is number of units produced,

So from above definition of marginal cost,

We get,

$MC=\dfrac{d\left( C(x) \right)}{dx}$………… (1)

Now in question it is given that the total cost which is,

$C(x)=0.007{{x}^{3}}-0.003{{x}^{2}}+15x+4000$

We need to find marginal cost when $17$ units produced i.e., Marginal cost($MC$) at

$x=17$,

Substituting this value in equation (1), we get

$MC=\dfrac{d\left( C(x) \right)}{dx}$

So substituting, we get,

$MC=\dfrac{d\left( 0.007{{x}^{3}}-0.003{{x}^{2}}+15x+4000 \right)}{dx}$

Differentiating separately we get,

\[MC=\dfrac{d\left( 0.007{{x}^{3}} \right)}{dx}-\dfrac{d\left( 0.003{{x}^{2}}

\right)}{dx}+\dfrac{d\left( 15x \right)}{dx}+\dfrac{d\left( 4000 \right)}{dx}\]

Taking constant out we get,

\[MC=(0.007)\dfrac{d\left( {{x}^{3}} \right)}{dx}-(0.003)\dfrac{d\left( {{x}^{2}}

\right)}{dx}+(15)\dfrac{d\left( x \right)}{dx}+\dfrac{d\left( 4000 \right)}{dx}\]

Now we know $\dfrac{d({{x}^{n}})}{dx}=n{{x}^{n-1}}$, so the above equation can be

differentiated as,

\[MC=(0.007)\left( 3{{x}^{2}} \right)-(0.003)\left( 2x \right)+(15)\left( 1 \right)+0\]

So simplifying in simple manner we get,

\[MC=0.021{{x}^{2}}-0.006x+(15)\]

\[MC=0.021{{x}^{2}}-0.006x+15\]…………… (2)

Now we have to find MC when $x=17$, So equation (2) can be written as,

\[MC=0.021{{(17)}^{2}}-0.006(17)+15\]

\[MC=0.021(289)-0.006(17)+15\]

So simplifying in simple manner,

\[MC=6.069-0.102+15\]

\[MC=20.967\]

Hence, required marginal cost i.e., $MC$ at $x=17$ is 20.967.


Note: You should be thorough with the concepts. You should know what marginal cost is. While solving the problem $MC=\dfrac{d\left( C(x) \right)}{dx}$ this identity is of utmost importance. While differentiating don’t get confused. While substituting the value of $x=17$ be careful. Sometimes while solving some terms are missed, take care of that.