Question

The cost, in dollars, of producing \[x\] gallons of detergent is given by $C\left( x \right) = 350 + 20x + 0.08{x^2} + 0.0004{x^3}$ . What is a formula for the marginal cost function $C'\left( x \right)$ ?
A) $20 + 0.16{x^2} + 0.0012{x^2}$
B) $20 + 0.16x + 0.0012{x^2}$
C) $20x + 0.16{x^2} + 0.0012{x^2}$
D) $20x + 0.16{x^2} + 0.0012{x^2}$

Answer 1

Hint: Observe that the cost function is expressed in terms of $x$. The marginal cost function is defined as the derivative of the function that expresses the cost of the gallons of detergent. We will differentiate the cost function as it is a simple polynomial function.

Complete step-by-step answer:
The cost of $x$ gallons of detergent is given by the polynomial function $C\left( x \right) = 350 + 20x + 0.08{x^2} + 0.0004{x^3}$.
The marginal cost is the change in the total cost of an object when the cost of production is increased per unit.
In the given example you will observe that the cost is depending on the variable $x$ , that means the marginal cost will be dependent on the same variable $x$ .
With respect to the given variable $x$ we will define the marginal cost as the rate of change of the cost with respect to change in the cost per unit.
Mathematically it indicates the rate of change in the function $C\left( x \right)$ with respect to the production of the units that is $x$ that is the derivative of the function $C\left( x \right)$ with respect to $x$ .
Now for a polynomial or algebraic function of the form $f\left( x \right) = {x^n}$ the derivative of the function with respect to $x$ is given by $f'\left( x \right) = n{x^{n - 1}}$ .
Therefore, we will differentiate the function $C\left( x \right)$with respect to $x$ and write the derivative as follows:
$C'\left( x \right) = 20 + 0.16x + 0.0012{x^2}$
Note that this derivative only denotes the marginal cost function for $x$ gallons of detergent.
Hence, the correct option is B.

Note: This is a good example of the applications of the derivative in real life. The function was given depending on one variable and on differentiating we get another function with the same variable. The interpretation of the derivative plays a vital role in the given problem.