A rectangle has an area 800 m^2. Length is twice of width. Length changes at 2 m/s and width at 1 m/s. What would be the rate of change when area becomes 40 m^2?
The length of the rectangle is twice the width. Initially the area is 800 m^2. If the width is equal to x, the length is 2x.
The area of the rectangle is 2x*x = 800
=> x^2 = 400
=> x = 20
The initial length is 40 m and the initial width is 20 m. The length decreases at the rate of 2 m/s and the width decreases at the rate of 1 m/s. After t seconds, the area of the rectangle is (40-2t)(20-t).
A = 800 - 80t - 2t^2
The rate of change of the area is -80 - 4t.
When the area of the rectangle is 40
40 = 800 - 80t - 2t^2
=> 380 - 40t - t^2 = 0
The positive root of the equation is `2*sqrt 195 - 20`
At this value of t , the rate at which the area is changing is `-80 - 4*(2sqrt 195 - 20)`
=> `-80 - 8*sqrt 195 + 80`
=> `-8*sqrt 195`
When the area of the rectangle is 40 m^2, its area is decreasing at `8*sqrt 195` m^2/s.