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 3.5 3.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 9 9 centimeters and the area is 83 83 square centimeters

Answer 1

-4 28/81 cm/min ≈ -4.346 cm/min

Explanation:

Differentiating the formula for the area of a triangle gives the relationship between the various rates of change.

A = 1/2bh

A' = 1/2(b'h +bh') . . . . . relation of rates of change

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triangle dimensions

When the area is 83 cm², and the height is 9 cm, the base length is ...

83 cm² = 1/2b(9 cm)

b = 166/9 cm

rate of change

Filling in the given values and rates of change, we have ...

3.5 cm²/min = 1/2(b'·(9 cm) +(166/9 cm)(2.5 cm/min))

7 cm²/min -46 1/9 cm²/min = (b')(9 cm)

b' = (-39 1/9 cm²/min)/(9 cm) = -4 28/81 cm/min

The base of the triangle is changing at the rate of -4 28/81 cm/min, about -4.346 cm/min.

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