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

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

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 (dA)/(dt)=(d)/(dt)\left[w\ell]`

By the Product Rule:

`\displaystyle (dA)/(dt)=(dw)/(dt)\ell +(d\ell)/(dt)w`

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

`\displaystyle (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 (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.