Question

The altitude (i.e., height) of a triangle is increasing at a rate of 2 cm/minute while the area of the triangle is increasing at a rate of 4.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 9.5 centimeters and the area is 92 square centimeters?

Answer 1

The formula for the area of a triangle is

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

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

We are given:

`(dh)/(dt)=2`

`(dA)/(dt)=4.5`

`h=9.5`

`A=92`

Plug in:

`4.5=(1)/(2)(b*2+9.5(db)/(dt))`
We can find
`b` with:

`A=(1)/(2)bh\rightarrowb=(2A)/(h)=(2*92)/(9.5)=(184)/(9.5)`
So
`4.5=(1)/(2)(2*(184)/(9.5)+9.5(db)/(dt))`

`\rightarrow9=(368)/(9.5)+9.5(db)/(dt)`

`\rightarrow9-(368)/(9.5)=9.5(db)/(dt)`

`\rightarrow(db)/(dt)=(9-(368)/(9.5))/(9.5)\approx-3.130`

So the base is decreasing at a rate of 3.130 cm/minute.