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 $2 cm^2 /min$. At what rate is the base of the triangle changing when the altitude is $10 cm$ and the area is $100 cm^2$.

Answer 1

let h is altitude & b is base of triangle
let db/dt = x
A = h*b/2
100 = 10*b/2
b = 20 cm.
dA/dt = 2
dh/dt = 1
dA/dt = (b*dh/dt + h*db/dt)/2
2 = (20*1 + 10*x)/2
x = -1.6
hence base of triangle is decreasing at the rate of -1.6cm²/min.