Question Coffee is dripping from a conical filter into a cylindrical coffeepot.

Question

Question

Coffee is dripping from a conical filter into a cylindrical coffeepot. Both the filter cone and the coffeepot have diameter of 6 inches. The cone has a height of 6 inches as well. When the coffee in the cone is 5 inches deep, the coffee is draining at the rate of 10in3/min.

(a). How fast is the level in the pot rising?

Hint: The volume of a cylinder of radius r and height h is V=?r2h.

(b). How fast is the level in the cone falling at the same time?

Hint: The volume of a cone of radius r and height h is V=1/3(?r2h)= ?r2h/3.

Question

(a). The profit P from selling x units of a product is given by P=15+12?x-18/x

(i). Find the average rate of change of P on the interval [1,9].

(ii). Find a formula for the marginal profit, and its numerical value when x=9

(b). A sphere of radius 3inches has its radius changing at a rate of 4 inches per second. How quickly is the volume of the sphere changing?

Explanation / Answer

The change in volume for the filter is -10 in^3/min
The change in volume for the coffee pot is +10 in^3/min

First, we need to describe the radius of the cone as a function of the height

When height = 6, diameter = 6 (radius = 3)
When height = 0, diameter = 0 (radius = 0)

So, h = 2r, or r = (1/2) * h

a)
V = pi * r^2 * h

Since the radius of the coffee pot is unchanging, we can treat it as a constant. r = 6

V = pi * 6^2 * h
V = 36 * pi * h

Derive:

dV/dt = 36 * pi * dh/dt

dV/dt = 10
Solve for dh/dt

10 = 36 * pi * dh/dt
5 = 18 * pi * dh/dt
dh/dt = 5 / (18 * pi)


b)
We want to find the rate at which the height is changing, so let's get the volume of the cone as a function of the height

V = (1/3) * pi * r^2 * h
r = (1/2) * h

V = (1/3) * pi * (1/2)^2 * h^2 * h
V = (1/3) * (1/4) * pi * h^3
V = (pi/12) * h^3

Now derive:

dV/dt = (pi/12) * 3 * h^2 * dh/dt
dV/dt = (pi/4) * h^2 * dh/dt

dV/dt = -10
h = 5
Solve for dh/dt

-10 = (pi/4) * 5^2 * dh/dt
-10 = (pi/4) * 25 * dh/dt
(-10/25) * (4/pi) = dh/dt
dh/dt = -40 / (25 * pi)
dh/dt = -8 / (5 * pi)

Since the question is asking how fast it is falling, then we can't say it's falling at -8/(5pi) inches per minute, but rather that it is falling at 8/(5pi) inches per minute.

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