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 1.5 1.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 8 8 centimeters and the area is 88 88 square centimeters

Answer 1

The base reduces at 3.75cm/min

Explanation:

Given

Let

`h \to altitude`

`b \to base`

`A \to Area`

So:

`(dh)/(dt) = 1.5cm/min`

`(dA)/(dt) = 1.5^2cm/min`

The area of a triangle is:

`A = (1)/(2)bh`

Calculate b when
`A =88cm^2; h =8cm`

`A = (1)/(2)bh`

`88=(1)/(2) * b * 8`

`88 =b * 4`

Solve for b

`b = 88/4`

`b = 22`

We have:

`A = (1)/(2)bh`

Differentiate with respect to time

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

Substitute the following values in the above equation

`(dh)/(dt) = 1.5cm/min`
`(dA)/(dt) = 1.5^2cm/min`
`b = 22`
`h = 8`

`1.5 = (1)/(2)(8 * (db)/(dt) + 22 * 1.5)`

Multiply both sides by 2

`3 = 8 * (db)/(dt) + 22 * 1.5`

`3 = 8 * (db)/(dt) + 33`

Collect like terms

`8 * (db)/(dt) = 3 -33`

`8 * (db)/(dt) = -30`

Divide both sides by 8

`(db)/(dt) = -(30)/(8)`

`(db)/(dt) = -3.75`