Answer:
The area of the rectangle increasing at the rate of 140 cm²/s
Step-by-step explanation:
Rectangle area:
A rectangle has two dimensions, length l and width w.
It's area is:
A = l*w.
When the length is 20 cm and the width is 10 cm, how fast is the area of the rectangle increasing?
We apply implicit differentiation to solve this question:
[tex]A = l*w[/tex]
So
[tex]\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}[/tex]
Length is 20, so [tex]l = 20[/tex].
Width is 10, so [tex]w = 10[/tex]
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.
This means that [tex]\frac{dl}{dt} = 8, \frac{dw}{dt} = 3[/tex]
So
[tex]\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}[/tex]
[tex]\frac{dA}{dt} = 20*3 + 10*8 = 140[/tex]
Area in cm².
So
The area of the rectangle increasing at the rate of 140 cm²/s
