Question

The altitude (i.e., height) of a triangle is increasing at a rate of 1.5 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 7.5 centimeters and the area is 80 square centimeters?

Answer 1

Rate at which base is decreasing is 3.07 cm /min

Explanation:

We have area of triangle,

                     A = 0.5 bh

Where b is base and h is altitude.

Differentiating with respect to time

               $A=0.5bh\\\\\frac{dA}{dt}=0.5\times \left ( b\times \frac{dh}{dt}+h\times \frac{db}{dt}\right )$

Here area is 80 square centimeters and altitude is 7.5 centimeters,

So we have

                 A = 0.5 bh

                 80 = 0.5 x b x 7.5

                   b = 21.33 cm

We also have

              $\frac{dh}{dt}=1.5cm/min\\\\\frac{dA}{dt}=4.5cm^2/min$

Substituting in differentiated equation

              $\frac{dA}{dt}=0.5\times \left ( b\times \frac{dh}{dt}+h\times \frac{db}{dt}\right )\\\\4.5=0.5\times \left ( 21.33\times 1.5+7.5\times \frac{db}{dt}\right )\\\\9= 32+7.5\times \frac{db}{dt}\\\\\frac{db}{dt}=-3.07cm/min$

Rate at which base is decreasing = 3.07 cm /min