Question

The altitude of a triangle is increasing at a rate of 2.5 2.5 centimeters/minute while the area of the triangle is increasing at a rate of 2 2 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7.5 7.5 centimeters and the area is 96 96 square centimeters

Answer 1

The base of the triangle is decreasing at a rate 8 centimeter per minute.

Explanation:

We are given the following in the question:

`(dh)/(dt) = 2.5\text{ cm per minute}\\\\(dA)/(dt) = 2\text{ square cm per minute}`

Instant height = 7.5 cm

Instant area = 96 square centimeters

Area of triangle =

`A=(1)/(2)* b* h`

where b is the base and h is the height of the triangle.

`96 = (1)/(2)* b * 7.5\\\\b = (96* 2)/(7.5) = 25.6`

Rate of change of area of triangle =

`(dA)/(dt) = (d)/(dt)((bh)/(2))\\\\(dA)/(dt) =(1)/(2)(b(dh)/(dt) + h(db)/(dt))`

Putting values, we get,

`2 = (1)/(2)(7.5(db)/(dt)+25.6(2.5))\\\\4 = 7.5(db)/(dt) + 64\\\\-60 = 7.5(db)/(dt) \\\\\Rightarrow (db)/(dt) = (-60)/(7.5) = -8`

Thus, the base of the triangle is decreasing at a rate 8 centimeter per minute.