Understanding Marginal Cost

The question asks us to find the marginal cost when 10 items are produced. The marginal cost represents the rate of change of the total cost with respect to the number of units produced. In calculus terms, it's the derivative of the total cost function, $C(x)$.

Total Cost Function

The total cost $C(x)$ in Rupees for producing $x$ units is given by the function:

$ C(x) = \text{0.007}x^3 + \text{26}x^2 + \text{15}x + \text{400} $

Calculating Marginal Cost

To find the marginal cost, we need to calculate the first derivative of the total cost function $C(x)$ with respect to $x$. Let's denote the marginal cost function as $C'(x)$.

$ C'(x) = \frac{dC}{dx} $

Now, we differentiate the given total cost function term by term:

• The derivative of $\text{0.007}x^3$ is $\text{0.007} \times 3x^{3-1} = \text{0.021}x^2$.
• The derivative of $\text{26}x^2$ is $\text{26} \times 2x^{2-1} = \text{52}x$.
• The derivative of $\text{15}x$ is $\text{15} \times 1x^{1-1} = \text{15}$.
• The derivative of the constant $\text{400}$ is 0.

So, the marginal cost function is:

$ C'(x) = \text{0.021}x^2 + \text{52}x + \text{15} $

Evaluating Marginal Cost at x=10

We need to find the marginal cost when 10 items are produced, which means we need to evaluate $C'(x)$ at $x = 10$.

Substitute $x = 10$ into the marginal cost function:

$ C'(10) = \text{0.021}(10)^2 + \text{52}(10) + \text{15} $

First, calculate $(10)^2 = 100$.

$ C'(10) = \text{0.021}(100) + \text{52}(10) + \text{15} $

Now, perform the multiplications:

• $\text{0.021} \times 100 = 2.1$
• $\text{52} \times 10 = 520$

Now, sum the terms:

$ C'(10) = 2.1 + 520 + 15 $

$ C'(10) = 537.1 $

The marginal cost when 10 items are produced is 537.1 Rupees.