CBSE Sample Paper 2024-25 · Section B · 2 marks

Q22: Marginal cost when 150 items are produced

The question

The cost (in rupees) of producing $x$ items in factory, each day is given by

\begin{aligned} C(x) &= 0.00013x^3 + 0.002x^2 \\ &\quad + 5x + 2200 \end{aligned}

Find the marginal cost when 150 items are produced.

Exam answer

The marginal cost function is

\begin{aligned} C'(x) &= 0.00039x^2 \\ &\quad + 0.004x + 5 \end{aligned}

Why?

Marginal cost is the rate of change of the total cost with respect to the number of items produced. So it is the derivative $C'(x)$ ⁠.

Differentiate term by term with the power rule:

\frac{d}{dx}(x^n) = nx^{n - 1}

\begin{aligned} 3 \times 0.00013 &= 0.00039 \\ 2 \times 0.002 &= 0.004 \end{aligned}

The term $5x$ gives 5, and the constant 2200 gives 0.

\begin{aligned} \frac{d}{dx}(0.00013x^3) &= 0.00013 \times 3x^2 \\ &= 0.00039x^2 \\[4pt] \frac{d}{dx}(0.002x^2) &= 0.002 \times 2x \\ &= 0.004x \\[4pt] \frac{d}{dx}(5x) &= 5 \times 1 = 5 \\ \frac{d}{dx}(2200) &= 0 \end{aligned}

Putting $x = 150$ ⁠,

\begin{aligned} C'(150) &= 0.00039(22500) \\ &\quad + 0.004(150) + 5 \\ &= 8.775 + 0.6 + 5 \\ &= 14.375 \end{aligned}

Why?

$150^2 = 22500$ ⁠. Then

\begin{aligned} 0.00039 \times 22500 &= 8.775 \\ 0.004 \times 150 &= 0.6 \end{aligned}

For the first product, $39 \times 225 = 8775$ ⁠, and the decimal point moves three places: $8.775$ ⁠.

\begin{aligned} 150^2 &= 150 \times 150 = 22500 \\[4pt] 0.00039 &= \frac{39}{100000} \\ 0.00039 \times 22500 &= \frac{39 \times 22500}{100000} \\ &= \frac{877500}{100000} = 8.775 \\[4pt] 0.004 \times 150 &= \frac{4 \times 150}{1000} \\ &= \frac{600}{1000} = 0.6 \end{aligned}

Now add, two at a time:

\begin{aligned} 8.775 + 0.6 &= 9.375 \\ 9.375 + 5 &= 14.375 \end{aligned}

The marginal cost is in rupees per item. It is the approximate cost of making one more item when 150 items are being made.

$\therefore$ The marginal cost when 150 items are produced is ₹ 14.375.

Where this comes from

Chapter 6, Application of Derivatives: Rate of change: marginal cost.

Question type: Marginal cost and marginal revenue

How to spot it: click to see

Type 3. Marginal cost and marginal revenue

Spot it: the words "marginal cost" or "marginal revenue", and a formula in $x$ ⁠, the number of units.

Here: The marginal cost is $C'(x) = 0.00039x^2 + 0.004x + 5$ ⁠. Put in the 150 items to get ₹ 14.375 per item.

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Application of Derivatives in past papers: question 23 of 37 · see them all