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 3.5 square cm/minute.

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 3.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 11 centimeters and the area is 88 square centimeters?

Answer 1

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 3.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 11 centimeters and the area is 88 square centimeters?

given: dA = 3.5 cm²/min

dh = 2.5 cm/min

A = 88 cm²

h = 11 cm

Equation for area of a triangle:

A = .5 b h

Plug in A and h to solve for b at that point:

A = .5 b h

88 = .5 (b) (11)

b = 88 / 5.5 = 16

Differentiate equation for area of a triangle to find rate of change of the area of a triangle (dA):

dA = .5(db)(h) + .5(b)(dh)

Plug in known variables to solve for the rate of change of the base (db):

dA = .5(db)(h) + .5(b)(dh)

3.5 = .5(db)(11) + .5(16)(2.5)

db = (3.5 - 20) / 5.5 = -3 cm/min