CA Foundation · Quantitative Aptitude · Differential and Integral Calculus
The total cost of producing x units is C(x) = 0.5x² + 20x + 800. The output at which the average cost per unit is minimum is:
Average cost is minimum at 40 units. Dividing cost by x gives AC = 0.5x + 20 + 800/x; setting its derivative 0.5 − 800/x² to zero gives x² = 1600, so x = 40, and the second derivative is positive there.
- Ax = 20
- Bx = 40Correct
- Cx = 80
- Dx = 16
Explanation
AC = 0.5x + 20 + 800/x. AC' = 0.5 − 800/x² = 0 gives x² = 1600, so x = 40. AC'' = 1600/x³ > 0, confirming a minimum. Choosing x = 20 would come from wrongly using x² = 400.
Did you get it right without looking?
One question tells you little. A timed set on Differential and Integral Calculus shows your real accuracy, how long you take and where you lose marks.
More Differential and Integral Calculus questions
- The value of lim (x→0) (e^(3x) − 1)/(5x) is:
- The area bounded by the curve y = x², the x-axis and the ordinates x = 1 and x = 3 is:
- The rate of change of revenue is R'(x) = 100 − 4x (in ₹), where x is units sold, and revenue is zero when no units are sold. The total reven…
- Evaluate ∫ (x + 5)/((x + 1)(x + 2)) dx (C is the constant of integration).
- If ∫ from 0 to k of (2x + 3) dx = 18, where k > 0, then the value of k is:
- The marginal cost of a firm is MC = 6x + 10 (in ₹), where x is output in units. If the fixed cost is ₹200, the total cost of producing 5 uni…