Question

The altitude (i.e., height) of a triangle is increasing at a rate of 2.5 cm/minute while the area of the triangle is increasing at a rate of 2.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 7 centimeters and the area is 84 square centimeters?

Answer 1

Answer:

The base is decreasing at a rate of 7.8571 centimeters per minute.

Step-by-step explanation:

We are given the following information in the question:

The height of a triangle is increasing at a rate of 2.5 cm/minute

$\displaystyle\frac{dh}{dt} = 2.5 \text{ cm per minute}$

The area of the triangle is increasing at a rate of 2.5 square cm/minute.

$\displaystyle\frac{dA}{dt} = 2.5 \text{ square cm per minute}$

Area of triangle is given by:

$A = \displaystyle\frac{1}{2}\times b\times h$

where A is the area of triangle, b is the base of triangle and h is the height of the triangle.

Differentiating, we get,

$A = \displaystyle\frac{1}{2}\times b\times h\\\\\frac{dA}{dt} = \frac{1}{2}\Bigg(b\frac{dh}{dt} + h\frac{db}{dt}\Bigg)$

We have to find rate of change of base of the triangle when the altitude is 7 centimeters and the area is 84 square centimeters

h = 7 cm

A  = 84 square centimeters

$84 = \displaystyle\frac{1}{2}\times 7\times b\\\\b = 24\text{ cm}$

Putting the values, we get:

$\displaystyle\frac{dA}{dt} = \frac{1}{2}\Bigg(b\frac{dh}{dt} + h\frac{db}{dt}\Bigg)\\\\2.5 = \frac{1}{2}\Bigg((24)(2.5) + (7)\frac{db}{dt}\Bigg)\\\\5 - 60 = 7\frac{db}{dt}\\\\\frac{dh}{dt} =-7.8571\text{ cm per minute}$

Thus, the base is decreasing at a rate of 7.8571 centimeters per minute.