At a sand and gravel plant, sand is falling off a conveyor and onto a conical pi

Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 10 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 15 feet high?

8/405pi feet per minute

[15pi] feet per minute

1/15pi feet per minute

{405pi/8] feet per minute

A manufacturer of desk lamps has determined that the cost to produce x units per month is given by

C=0.2x2+10,000.

How fast is cost per month changing when production is changing at the rate of 12 units per month and the production level is 80 units?

$11280 per month

$320 per month

$384 per month

$192 per month

Explanation / Answer

I.

First we need a formula to relate the radius and height of a cone. Volume will do nicely, especially considering they gave you the rate volume is changing.

V = (1/3)R²h

Now since we do not have a value for dR/dt, we will need to put R in terms of h. Luckily they gave us a proportion. The diameter is 2R.

2R = 3h
R = 3h/2

Now substitute this back into the formula.

V = (1/3)(3h/2)²h

Simplify a little to make your life easier.

V = (1/3)(9h²/4)h

V = (3/4)h³

Now just derive (in terms of time) and plug in 15 for the height and 10 for the change in volume, then solve for the change in height.

dV/dt = (3/4)(3h²)(dh/dt)

dV/dt = 10 ft³/min
h = 15 ft.

10 = (3/4)(3(15)²)(dh/dt) -- plug in given values

10 = (2025 / 4)(dh/dt) -- simplify

40 / 2025 ft. / min= dh/dt -- solve

dh/dt = 8 / 405 ft. / min

II.

Given A manufacturer of desk lamps has determined that the cost to produce x units per month is given by

C = 0.2x^2+10,000

cost per month changing when production is changing at the rate of 12 units per month and the production level is 80 units

dC / dt = (dC / dx )*(dx / dt)

=0.2*2*x*(X)

= 0.4*12*80 = 384 (therefore x=12, X=80))

$384 per month

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