The altitude of a triangle is increasing at a rate of 1.000 centimeters/minute while the area of the triangle is increasing at a rate of 3.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 11.500 centimeters and the area is 85.000 square centimeters?

Note: The "altitude" is the "height" of the triangle in the formula "Area=(1/2)*base*height". Draw yourself a general "representative" triangle and label the base one variable and the altitude (height) another variable. Note that to solve this problem you don't need to know how big nor what shape the triangle really is.

(1/2) b h = A

(1/2) b (11.5) = 85
b = 14.8 if we need it

2 dA/dt = b dh/dt + h db/dt

2 (3) = 14.8 (1) + 11.5 db/dt
- 8.8 = 11.5 db/dt
db/dt = - .765

Water is leaking out of an inverted conical tank at a rate of 10600.0 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 14.0 m and the the diameter at the top is 6.0 m. If the water level is rising at a rate of 27.0 cm/min when the height of the water is 1.5 m, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

To find the rate at which the base of the triangle is changing, we need to use the related rates concept from calculus.

Let's denote the altitude of the triangle as "h", the base as "b", and the area as "A". We are given that dh/dt (rate at which altitude is changing) = 1.000 cm/min and dA/dt (rate at which area is changing) = 3.000 cm²/min.

We need to find db/dt (rate at which the base is changing) when h = 11.500 cm and A = 85.000 cm².

We know that the area of a triangle is given by the formula A = (1/2) * b * h, where b is the base and h is the altitude. We can differentiate this equation with respect to time (t) using implicit differentiation:

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

Now we can substitute the known values to solve for db/dt:

3.000 = (1/2) * db/dt * 11.500 + (1/2) * b * 1.000

To isolate db/dt, we move terms around:

(1/2) * db/dt * 11.500 = 3.000 - (1/2) * b * 1.000

db/dt * 11.500 = 6.000 - 0.500 * b

db/dt = (6.000 - 0.500 * b) / 11.500

Now, we can substitute the given values h = 11.500 cm and A = 85.000 cm²:

db/dt = (6.000 - 0.500 * 85.000) / 11.500

Simplifying the equation:

db/dt = (6.000 - 42.500) / 11.500

db/dt = -36.500 / 11.500

db/dt ≈ -3.174 cm/min

Therefore, the rate at which the base of the triangle is changing when the altitude is 11.500 cm and the area is 85.000 cm² is approximately -3.174 cm/min. Note that the negative sign indicates the base is decreasing.