# 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?

justaguide | College Teacher | (Level 2) Distinguished Educator

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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.