Question

The altitude of a triangle is increasing at a rate of 1.5 cm/min while the area of the triangle is increasing at a rate of 5 square cm/min. At what rate is the base of the triangle changing when the altitude is 9 cm and the area is 81 square cm?

Answer 1

-1 8/9 cm/min

Explanation:

The area of a triangle is given by ...

A = (1/2)bh

Then when the area is 81 cm² and the altitude is 9 cm, the base is ...

81 cm² = (1/2)(b)(9 cm)

(81 cm²)/(4.5 cm) = b = 18 cm

Differentiating the area formula implicitly, you have ...

A' = (1/2)(b'h +bh')

Fill in the given values and solve for b':

5 cm²/min = (1/2)(b'(9 cm) +(18 cm)(1.5 cm/min))

5 cm²/min = b'(4.5 cm) +13.5 cm²/min . . . . simplify

Subtract 13.5 and divide by the coefficient of b':

(-8.5 cm²/min)/(4.5 cm) = b' = -1 8/9 cm/min

The base is changing at the rate of -1 8/9 ≈ -1.889 cm per minute.