Question

The length of a rectangle is increasing at a rate of 6 cm/s and its width is increasing at a rate of 5 cm/s. When
the length is 12 cm and the width is 4 cm, how fast is the area of the rectangle increasing (in cm/s)? Write an equation for A in terms of l and w.

Answer 1

Answer:

The area of the rectangle is increasing at a rate of 84 square centimeters per second.

Step-by-step explanation:

The area for a rectangle is given by the formula:

$A=w\ell$

Where w is the width and l is the length.

We are given that the length of the rectangle is increasing at a rate of 6 cm/s and that the width is increasing at a rate of 5 cm/s. In other words, dl/dt = 6 and dw/dt = 5.

First, differentiate the equation with respect to t, where w and l are both functions of t:

$\displaystyle \frac{dA}{dt}=\frac{d}{dt}\left[w\ell]$

By the Product Rule:

$\displaystyle \frac{dA}{dt}=\frac{dw}{dt}\ell +\frac{d\ell}{dt}w$

Since we know that dl/dt = 6 and that dw/dt = 5:

$\displaystyle \frac{dA}{dt}=5\ell + 6w$

We want to find the rate at which the area is increasing when the length is 12 cm and the width is 4 cm. Substitute:

$\displaystyle \frac{dA}{dt}=5(12)+6(4)=84\text{ cm}^2\text{/s}$

The area of the rectangle is increasing at a rate of 84 square centimeters per second.