The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm2/min.

Question

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm²/min.

 At what rate is the base of the triangle changing when the altitude is 10 cm

 and the area is 100 cm²?

Answer 1

A= bh/2; dA/dt = h(db/dt)/2 + b(dh/dt)/2.

So, if A = 100 cm² h = 10 cm, then b = 20 cm and 2 = 10(db/dt)/2 + 20(1)/2; db/dt = - 1.6 cm/min