the length of a rectangle is increasing at a rate of 3 cm/s and its width is increasing at a rate of 8 cm/s…

the length of a rectangle is increasing at a rate of 3 cm/s and its width is increasing at a rate of 8 cm/s. when the length is 11 cm and the width is 4 cm, how fast is the area of the rectangle increasing (in cm²/s)?

the length of a rectangle is increasing at a rate of 3 cm/s and its width is increasing at a rate of 8 cm/s. when the length is 11 cm and the width is 4 cm, how fast is the area of the rectangle increasing (in cm²/s)?

Answer

Explanation:

Step1: Write the area formula for a rectangle

The area formula for a rectangle is (A = lw), where (l) is the length and (w) is the width.

Step2: Differentiate the area formula with respect to time (t)

Using the product rule ((uv)^\prime=u^\prime v + uv^\prime), where (u = l) and (v = w). (\frac{dA}{dt}=\frac{dl}{dt}w + l\frac{dw}{dt})

Step3: Substitute the given values

We are given that (\frac{dl}{dt}=3) cm/s, (\frac{dw}{dt}=8) cm/s, (l = 11) cm, and (w = 4) cm. (\frac{dA}{dt}=(3)(4)+(11)(8)) (\frac{dA}{dt}=12 + 88)

Answer:

(100)