Question

The length of a rectangle is increasing at a rate of 3 cm/s and its width is increasing at a rate of 4 cm/s. When the length is 15 cm and the width is 11 cm, how fast is the area of the rectangle increasing

Answer 1

The rectangle's area is the product of its length and width. Differentiating both the area and the product with respect to time $t$ gives

$A=\ell w\implies\dfrac{\mathrm dA}{\mathrm dt}=\ell\dfrac{\mathrm dw}{\mathrm dt}+\dfrac{\mathrm d\ell}{\mathrm dt}w$

The length and width increase at respective rates of $\frac{\mathrm d\ell}{\mathrm dt}=3\frac{\rm cm}{\rm s}$ and $\frac{\mathrm dw}{\mathrm dt}=4\frac{\rm cm}{\rm s}$.

At the moment $\ell=15\,\mathrm{cm}$ and $w=11\,\mathrm{cm}$, the area is increasing at a rate of

$\dfrac{\mathrm dA}{\mathrm dt}=(15\,\mathrm{cm})\left(3\dfrac{\rm cm}{\rm s}\right)+\left(4\dfrac{\rm cm}{\rm s}\right)(11\,\mathrm{cm})=89\dfrac{\mathrm{cm}^2}{\rm s}$