Question

The altitude of a triangle is increasing at a rate of 1.5 1.5 centimeters/minute while the area of the triangle is increasing at a rate of 2.5 2.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 8.5 8.5 centimeters and the area is 93 93 square centimeters?

Answer 1

The base is decreasing at a rate 3.27 centimeters per minute.

Explanation:

We are given the following in the question:

`(dh)/(dt)= 1.5\text{ centimeters per minute}\\\\(dA)/(dt) = 2.5\text{ square centimeters per minute}`

Instant altitude = 8.5 cm

Area = 93 square centimeters

Area of triangle:

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

Putting values, we get,

`93 = (1)/(2)* b* 8.5\\\\b = (93* 2)/(8.5)`

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

Rate of change of area =

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

Putting values, we get,

`2.5 = (1)/(2)(8.5(db)/(dt) + (186)/(8.5)* 1.5)\\\\5 = 8.5(db)/(dt) + (186)/(8.5)* 1.5\\\\8.5(db)/(dt) = 5 - (186)/(8.5)* 1.5 \\\\8.5(db)/(dt) = -27.82\\(db)/(dt) = -3.27`

Thus, the base is decreasing at a rate 3.27 centimeters per minute.