The length of a rectangle falls at 3 cm/min while its width rises at 2 cm/min. When the length is 8 cm and the width 6 cm, how fast is the area changing?

Explanation

1. 01

   $A = lw$, so $\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$.

2. 02

   $8 \times 2 + 6 \times (-3) = 16 - 18 = -2$: the area is shrinking at 2 cm²/min.