Question

1. the length of a rectangle is increasing at a rate of 8 cm/s and its width is increasing at a rate of 3 cm/s. when the length is 20 cm and the width is 10 cm, how fast is the area of the rectangle increasing?

Explanation:

Step1: Recall the area formula

Let the length of the rectangle be $l$ and the width be $w$. The area $A = lw$.

Step2: Differentiate with respect to time $t$

Using the product - rule $\frac{d(uv)}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$, we have $\frac{dA}{dt}=l\frac{dw}{dt}+w\frac{dl}{dt}$.

Step3: Substitute the given values

We know that $\frac{dl}{dt}=8$ cm/s, $\frac{dw}{dt}=3$ cm/s, $l = 20$ cm and $w = 10$ cm.
Substituting these values into the formula for $\frac{dA}{dt}$, we get $\frac{dA}{dt}=20\times3 + 10\times8$.
$\frac{dA}{dt}=60 + 80$.

Answer:

$140$ cm²/s