The length \( x \) of a rectangle is decreasing at the rate of 3 cm/min and the width \( y \) is increasing at the rate of 2 cm/min. If \( x=10 \) cm and \( y=6 \) cm, then the rate of change (increase/decrease) of the area of the rectangle is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2 cm2/min
The area is A = xy. Since x is decreasing, its rate carries a minus sign:
\( \displaystyle \frac{dx}{dt}=-3\ \text{cm/min} \)
\( \displaystyle \frac{dy}{dt}=2\ \text{cm/min} \)
By the product rule,
\( \displaystyle \frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt} \)
\( \displaystyle =10(2)+6(-3) \)
\( \displaystyle =20-18=2\ \text{cm}^2\text{/min} \)
The value is positive, so the area is increasing at 2 cm2/min.
Why the other options are incorrect
Option (1) 20 cm2/min — Incorrect. 20 is only the term \( x\frac{dy}{dt} \); the change in length also affects the area.
Option (2) 38 cm2/min — Incorrect. 38 = 20 + 18 treats the length as increasing; since x is decreasing, \( \frac{dx}{dt}=-3 \).
Option (4) 18 cm2/min — Incorrect. 18 is only the size of the term \( y\frac{dx}{dt} \); the change in width also affects the area.
NCERT (2026-27 edition):
Rate of Change of Quantities
