#### To determine

The area of the rectangle is increasing when the length is 20 cm and the width is 10 cm.

#### Answer

The area of rectangle is increasing at the rate of dAdt=140 cm2\s.

#### Explanation

**Derivative rules:**

(1) *Product rule* ddx(fg)=fddx(g)+gddx(f)

(2) *Chain rule*: dydx=dydu⋅dudx

**Calculation:**

Let *A* be an area of the rectangle with length of the rectangle *l* and width of the rectangle is *w*.

The length and width are increasing when the time *t* is increasing. That is, the length and width are depends on time variable *t*. Thus, the length *l* and the width *w* are a function of the variable *t*.

Note that, the rates of change is a derivatives.

The length of the rectangle is increasing at a rate of 8 cm\s and with is increasing at a rate of 3 cm\s. That is, dldt=8 cm/s and dwdt=3 cm/s.

Obtain the derivatives dAdt when length is 20 cm and width 10 cm of the rectangle.

The area of the rectangle is, A=l×w.

Differentiate with respect to time variable *t*.

dAdt=ddt(l×w)

Apply the product rule (1) and the chain rule (2),

dAdt=lddt(w)+wddt(l)=l[ddw(w)dwdt]+w[ddl(l)dldt]=l[(1)dwdt]+w[(1)dldt]=ldwdt+wdldt

Substitute the value l=20 cm , w=10 cm, dldt=8 cm/s and dwdt=3 cm/s in dAdt.

dAdt=(20 cm)(3 cm\s)+(10 cm)(8 cm\s)=(60+80)cm2\s=140 cm2\s

Therefore, the area of rectangle is increasing at the rate of dAdt=140 cm2\s.