The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm²/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm²?

Solution:

Given

dh/dt = 1 cm/min

dA/dt = 2 cm²/min

When A = 100 cm², h = 10 cm

We know that

A = 1/2 × b × h

b = 20 cm

Let us differentiate both sides

dA/dt = 1/2 [b dh/dt + h db/dt]

Substituting the values

2 = 1/2 [20 × 1 + 10 × db/dt]

By further simplification

4 = 20 + 10 db/dt

db/dt = -1.6 cm/min

Therefore, the base of the triangle is decreasing at the rate of 1.6 cm/min.

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The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm²/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm²?

Summary:

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm²/min. The base of the triangle is decreasing at the rate of 1.6 cm/min when the altitude is 10 cm and the area is 100 cm²