Question

The altitude of a triangle is increasing at a rate of 2.5 2.5 centimeters/minute while the area of the triangle is increasing at a rate of 2 2 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7.5 7.5 centimeters and the area is 96 96 square centimeters

Answer 1

Answer:

The base of the triangle is decreasing at a rate 8 centimeter per minute.      

Step-by-step explanation:

We are given the following in the question:

$\dfrac{dh}{dt} = 2.5\text{ cm per minute}\\\\\dfrac{dA}{dt} = 2\text{ square cm per minute}$

Instant height = 7.5 cm

Instant area = 96 square centimeters

Area of triangle =

$A=\dfrac{1}{2}\times b\times h$

where b is the base and h is the height of the triangle.

$96 = \dfrac{1}{2}\times b \times 7.5\\\\b = \dfrac{96\times 2}{7.5} = 25.6$

Rate of change of area of triangle =

$\dfrac{dA}{dt} = \dfrac{d}{dt}(\dfrac{bh}{2})\\\\\dfrac{dA}{dt} =\dfrac{1}{2}(b\dfrac{dh}{dt} + h\dfrac{db}{dt})$

Putting values, we get,

$2 = \dfrac{1}{2}(7.5\dfrac{db}{dt}+25.6(2.5))\\\\4 = 7.5\dfrac{db}{dt} + 64\\\\-60 = 7.5\dfrac{db}{dt} \\\\\Rightarrow \dfrac{db}{dt} = \dfrac{-60}{7.5} = -8$

Thus, the base of the triangle is decreasing at a rate 8 centimeter per minute.