The area of an expanding rectangle is increasing at the rate of 48 cm^2 / s. The length of the rectangle is always square of its breadth. At what…

Let the length and breadth of the expanding rectangle at any time be ' $\displaystyle x$ ' and ' $\displaystyle y$ ' respectively. Then, $\displaystyle x=y^{2}, A($ Area $\displaystyle )=x y=x^{\frac{3}{2}}$ \[\frac{d A}{d t}=\frac{3}{2} \sqrt{x} \frac{d x}{d t} \] $\displaystyle \Rightarrow 48=\frac{3}{2} \sqrt{(4.5)^{2}} \frac{d x}{d t} \Rightarrow \frac{d x}{d t}=\frac{64}{9} \mathrm{~cm} / \mathrm{s}$