Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. how fast is the height of the pile increasing when the pile is 14 ft high? (round your answer to two decimal places.)

This involves relationship between rates using Calculus.

dh/dt = 0.13 ft/min

Volumetric rate; dv/dt = 20 ft³/min

height of pile; h = 14 ft

We are not given the diameter here but as we are dealing with a right circular cone, we will assume that the diameter is equal to the height.

Thus; diameter; d = 14 ft

radius; r = h/2 = d/2 = 14/2

radius; r= 7 ft

Plugging in the relevant values, we have;

20 = ¹/₄π × 14² × (dh/dt)

dh/dt = (20 × 4)/(π × 14²)

dh/dt = 0.13 ft/min

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The height of the pile is increasing a minute when the pile is 14ft high.

Step-by-step explanation:

The volume of a cone is given by the following formula:

We have that the diameter and the height are equal, so

So

Let's derivate this equation, using implicit derivatives.

In this problem, we have to:

Find , when . So

The height of the pile is increasing a minute when the pile is 14ft high.

total of apples - take away four

6-4 = 2

answer is 2 apples