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
$A=\frac{1}{2}bh$
$\rightarrow\frac{dA}{dt}=\frac{1}{2}(b\frac{dh}{dt}+h\frac{db}{dt})$

We are given:
$\frac{dh}{dt}=2$
$\frac{dA}{dt}=4.5$
$h=9.5$
$A=92$

Plug in:
$4.5=\frac{1}{2}(b*2+9.5\frac{db}{dt})$
We can find $b$ with:
$A=\frac{1}{2}bh\rightarrowb=\frac{2A}{h}=\frac{2*92}{9.5}=\frac{184}{9.5}$
So $4.5=\frac{1}{2}(2*\frac{184}{9.5}+9.5\frac{db}{dt})$
$\rightarrow9=\frac{368}{9.5}+9.5\frac{db}{dt}$
$\rightarrow9-\frac{368}{9.5}=9.5\frac{db}{dt}$
$\rightarrow\frac{db}{dt}=\frac{9-\frac{368}{9.5}}{9.5}\approx-3.130$

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