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 determine the marginal cost when 70 geometry boxes are produced, we need to analyze the cost components involved as mentioned: 

• Raw Material Cost: For 'x' units, each unit's cost is ₹(x² + 2). Total raw material cost for 'x' units is \(x(x^2 + 2) = x^3 + 2x\).
• Transportation Cost: Given as half the number of boxes produced, it is \( \frac{x}{2} \).
• Storage Cost: Fixed cost of ₹150 per day.

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

To find the marginal cost, compute the derivative of the cost function \( C(x) \), denoted as \( C'(x) \). The expression for \( C(x) \) simplifies as follows:

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

Differentiate \( C(x) \) with respect to 'x':

• \( C'(x) = \frac{d}{dx}(x^3 + 2x + \frac{x}{2} + 150) = 3x^2 + 2 + \frac{1}{2} \)
• \( C'(x) = 3x^2 + 2x + \frac{1}{2} \)
• Simplify: \( C'(x) = 3x^2 + 2.5 \)
• Note the transportation term correct factor: \( C'(x) = 3x^2 + 2.5 \) checks the fixed computation.

Evaluate \( C'(x) \) at \( x = 70 \):

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

Thus, the marginal cost when 70 geometry boxes are produced is ₹14,702.50.

Answer 2

The total cost $C(x)$ is the sum of:
Raw material cost: $x(x^2 + 2) = x^3 + 2x$,
Transportation cost: $\frac{5x}{2}$,
Storage cost: 150.
Thus:
$C(x) = x^3 + 2x + \frac{5x}{2} + 150$.
Simplify:
$C(x) = x^3 + \frac{9x}{2} + 150$.
Step 1: Find the marginal cost.
The marginal cost is the derivative of $C(x)$:
$C'(x) = \frac{d}{dx} \left(x^3 + \frac{9x}{2} + 150\right)$.
Differentiate term by term:
$C'(x) = 3x^2 + \frac{9}{2}$.
Step 2: Evaluate at $x = 70$.
Substitute $x = 70$ into $C'(x)$:
$C'(70) = 3(70)^2 + \frac{9}{2}$.
Simplify:
$C'(70) = 3(4900) + \frac{9}{2} = 14700 + 4.5 = 14702.5$.
Final Answer:
14,702.5