Basic Applications of CalculusPYQ Jan 26Question 4517 of 28
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The cost function of a company is given by C(x)=600x−10x2+x32\displaystyle C(x) = 600x - 10x^2 + \frac{x^3}{2} where x denotes the output. Find the level of output (in nearest integer) at which average cost is minimum.

Options

Ax=8\displaystyle x = 8
Bx=9\displaystyle x = 9
Cx=10\displaystyle x = 10
Dx=11\displaystyle x = 11
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Correct Answer

✅ Option c — x=10\displaystyle x = 10

All Options:

  • Ax=8\displaystyle x = 8
  • Bx=9\displaystyle x = 9
  • Cx=10\displaystyle x = 10
  • Dx=11\displaystyle x = 11

Detailed Solution & Explanation

Given the cost function:
C(x)=600x−10x2+x32C(x) = 600x - 10x^2 + \frac{x^3}{2}
where x\displaystyle x represents the level of output.

**Step 1: Find the Average Cost (AC\displaystyle AC) function**
AC=C(x)x=600−10x+x22AC = \frac{C(x)}{x} = 600 - 10x + \frac{x^2}{2}

**Step 2: Find the first derivative of AC\displaystyle AC with respect to x\displaystyle x**
d(AC)dx=ddx(600−10x+x22)\frac{d(AC)}{dx} = \frac{d}{dx}\left(600 - 10x + \frac{x^2}{2}\right)d(AC)dx=−10+x\frac{d(AC)}{dx} = -10 + x

**Step 3: Set the first derivative to 0\displaystyle 0 to find the critical points**
−10+x=0  ⟹  x=10-10 + x = 0 \implies x = 10

**Step 4: Check the second derivative for minimality**
d2(AC)dx2=ddx(−10+x)=1\frac{d^2(AC)}{dx^2} = \frac{d}{dx}(-10 + x) = 1
Since the second derivative is positive (1>0\displaystyle 1 > 0), the average cost is minimized at x=10\displaystyle x = 10.

Hence, **Option C** is the correct answer.

About This Chapter: Basic Applications of Calculus

Paper

Paper 3: Quantitative Aptitude

Weightage

3-5 Marks

Key Topics

Limits, Continuity, Derivatives, Integrals

This chapter covers Limits, Continuity, Derivatives, Integrals and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 3-5 Marks weightage. Focus on understanding core concepts rather than memorizing.

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