how-fast-is-the-area-of-a-rectangle-changing-if-one-side-is-10-cm-long-and-increasing-at-a-rate-of-2-cm-s-and-the-other-side-is-8-cm-long-and-is-decreasing-at-a-rate-of-3-cm-s-

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Question Number 132211 by benjo_mathlover last updated on 12/Feb/21

$$$$$$\mathrm{how}\mathrm{fast}\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{a}$$$$\mathrm{rectangle}\mathrm{changing}\mathrm{if}\mathrm{one}\mathrm{side}\mathrm{is}10$$$$\mathrm{cm}\mathrm{long}\mathrm{and}\mathrm{increasing}\mathrm{at}\mathrm{a}$$$$\mathrm{rate}\mathrm{of}2\mathrm{cm}/\mathrm{s}\mathrm{and}\mathrm{the}\mathrm{other}\mathrm{side}\mathrm{is}$$$$8\mathrm{cm}\mathrm{long}\mathrm{and}\mathrm{is}\mathrm{decreasing}\mathrm{at}$$$$\mathrm{a}\mathrm{rate}\mathrm{of}3\mathrm{cm}/\mathrm{s}$$

Answered by ajfour last updated on 12/Feb/21

$$A=xy$$$$\frac{dA}{dt}=y\left(\frac{dx}{dt}\right)+x\left(\frac{dy}{dt}\right)$$$$=\left(8cm\right)\left(\frac{2cm}{s}\right)+\left(10cm\right)(-\frac{3cm}{s})$$$$\Rightarrow \frac{dA}{dt}=-\frac{14{cm}^2}{s}$$$$\Rightarrow Areaisdecreasingatarate$$$$of\frac{14{cm}^2}{s}.$$

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