# the altitude of a triangle is increasing at a rate of 2.500 centimeters/ minute while the area of a triangle is increasing at a rate of 3.000 square centimeters/minute. At what rate is the base of...

the altitude of a triangle is increasing at a rate of 2.500 centimeters/ minute while the area of a triangle is increasing at a rate of 3.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 10.000 centimeters and the area is 90.000 square centimeters?

*print*Print*list*Cite

The area of a triangle is given by `A = (1/2)*b*h` where b is the base and h is the altitude.

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

The altitude of the triangle is increasing at a rate of 2.5 centimeters/ minute while the area of a triangle is increasing at a rate of 3.0 square centimeters/minute.

When the altitude is 10.0 centimeters and the area is 90.0 square centimeters, the base is 18. This gives:

`3 = (1/2)*18*2.5 + (1/2)*10*((db)/dt)`

=> `(db)/dt = -3.9`

**The length of the base of the triangle is decreasing at the rate 3.9 cm/minute**

Area of triangle=A

Base =b

Altitude =l

Thus

`A=(1/2)bl` (i)

90=(1/2)bx10

b=18cm.

differentiate (i) w.r.t. 't' ,we have

`(dA)/(dt)=(1/2)((db)/(dt)xxl+bxx(dl)/(dt))`

`(dA)/(dt)=+3.0 (cm)/min.,`

`(dl)/(dt)=+2.5 (cm)/(min.)`

`3.0=(1/2)((db)/(dt)xx10+18xx(2.5))`

`6-45=(db)/(dt)xx10`

`-39/10 (cm)/(min.)=(db)/(dt)`

`(db)/(dt)=-3.9 (cm) / min`