Question

the altitude of a triangle is increasing at a rate of 1 cm/min while the area in 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 100cm^2?

Answer 1

I suppose \[
A = \frac 12 bh
\]And thus by power rule \[
\frac {dA}{dt}=\frac 12\frac{db}{dt}h+\frac 12 b\frac {dh}{dt}
\]

Answer 2

Area of Triangle is
\[A = \frac{bh}{2}\]

If we take the derivative, we get:

\[A'(t) = \frac{b'h + bh'}{2}\]

After inputting the given values we have:

\[100 = \frac{b(10)}{2}\]
\[b = 20\]

And

\[2 = \frac{b'(10) + 20(1)}{2}\]

Solving for \(b'\) we get:
\[b' = \frac{4 - 20}{10} = -\frac{8}{5}\]

Therefore the rate at which the base of the triangle is changing is -8/5 cm/min