Question

The altitude of a triangle is increasing at a rate of 2.5 cm/min while the area of the triangle is increasing at a rate of 1.5 square cm/min. At what rate is the base of the triangle changing when the altitude is 11 cm and the area is 97 square cm?

cm/min

Answer 1

Explanation:

We know that the area A of a triangle is given by the formula A = 1/2 * base * height. Let's denote the base as b and the height (altitude) as h. The rates are given as dA/dt = 1.5 square cm/min and dh/dt = 2.5 cm/min. We're asked to find db/dt, the rate of change of the base, at a specific point in time when h = 11 cm and A = 97 square cm.

We can start by differentiating the area formula with respect to time t, which gives us:

dA/dt = 1/2 * (b * dh/dt + h * db/dt).

We know the values of dA/dt, dh/dt, h, and A, and we can solve for b using the area formula:

b = 2A/h = 2*97/11 ≈ 17.64 cm.

Substituting all the known values into the derivative equation:

1.5 = 1/2 * (17.64 * 2.5 + 11 * db/dt).

Solving this equation for db/dt gives us:

db/dt = (1.52 - 17.642.5) / 11 ≈ -1.77 cm/min.

So, the base of the triangle is decreasing at a rate of approximately 1.77 cm per minute at the instant when the height is 11 cm and the area is 97 square cm.