Question

A company produces 'x' units of geometry boxes in a day. If the raw material of one geometry box costs ₹2 more than the square of the number of boxes produced in a day, the cost of transportation is half the number of boxes produced in a day, and the cost incurred on storage is ₹150 per day. The marginal cost (in ₹) when 70 geometry boxes are produced in a day is:

• A. ₹14,852.50
• B. ₹14,702.50
• C. ₹14,795
• D. ₹5,087.50

Answer 1

The Correct Option is B

To ascertain the marginal cost for producing 70 geometry boxes, the following cost components must be analyzed:

• Raw Material Cost: For 'x' units, the cost per unit is ₹(x² + 2). The total raw material cost for 'x' units is \(x(x^2 + 2) = x^3 + 2x\).
• Transportation Cost: This is calculated as half the number of boxes produced, represented by \( \frac{x}{2} \).
• Storage Cost: A fixed daily cost of ₹150.

The aggregate cost function for producing 'x' units is \( C(x) = x^3 + 2x + \frac{x}{2} + 150 \).

The marginal cost is derived by computing the derivative of the cost function \( C(x) \), denoted as \( C'(x) \). The cost function \( C(x) \) is simplified as:

• \( C(x) = x^3 + 2x + \frac{x}{2} + 150 \)
• Consolidating terms: \( C(x) = x^3 + \frac{4x}{2} + 150 = x^3 + 2x + 150 \)

Differentiating \( C(x) \) with respect to 'x' yields:

• \( C'(x) = \frac{d}{dx}(x^3 + 2x + \frac{x}{2} + 150) = 3x^2 + 2 + \frac{1}{2} \)
• \( C'(x) = 3x^2 + 2x + \frac{1}{2} \)
• Simplified: \( C'(x) = 3x^2 + 2.5 \)
• Verification of the fixed computation with the transportation term: \( C'(x) = 3x^2 + 2.5 \).

The derivative is evaluated at \( x = 70 \):

• \( C'(70) = 3(70)^2 + 2.5 \)
• \( C'(70) = 3(4900) + 2.5 \)
• \( = 14700 + 2.5 = 14702.5 \)

Therefore, the marginal cost for producing 70 geometry boxes is ₹14,702.50.