The hopper bin in the picture below is modeled by the two-dimensional figure beside it. The hopper bin is
composed of a cylinder, and 2 EQUAL cones. Let d be the center of the base of the cone.
52° cylinder cones
The diameter of the hopper is 36 ft and the height of the cylindrical portion is 20 ft. The measure of the acute
angle formed by the slant height of the cone and its base is 52°, determine and state, to the nearest cubic foot,
the volume of the hopper bin.
The hopper bin was constructed to hold a maximum of 500,000 pounds of feed. If feed weighs 22 pounds per cubic foot, can the carnival tent be filled to 85% of its volume and not exceed the weight limit? Justify your
answer.

The volume of the hopper bin is 35974 cubic feet
The carnival tent can be filled to 85% of its volume and not exceed the weight limit

The given parameters are:

Start by calculating the radius (r) of the cones.

r = D/2 -------- because the cones and the cylinder have equal diameters.

So, we have:

r = 36/2 = 18

The height (h) of the cones is:

tan(θ) = h/r

This gives

tan(52) = h/18

Make h the subject

h = 18 * tan(52)

Evaluate

h = 23.04

The volume is then calculated as:

Volume = Volume of cylinder + 2 * Volume of cone

This gives

Volume = πr^2H + 2 * 1/3πr^2h

Volume = πr^2H + 2/3πr^2h

Factor out πr^2

Volume = πr^2(H + 2/3h)

Substitute known values

Volume = 3.14 * 18^2 * (20 + 2/3 * 23.04)

Evaluate

Volume = 35973.85 cubic feet

Hence, the volume of the hopper bin is 35974 cubic feet

The given parameters are:

For a feed that weighs 22 pounds per cubic feet, the weight of the hopper bin would be:

Weight = 22 * 35974 pounds

Weight = 791428 pounds

At 85% full, the weight wuld be:

Weight = 85% * 791428 pounds

Weight = 672713.8 pounds

672713.8 pounds is greater than the weight limit of 500000 pounds.

Hence, the carnival tent can be filled to 85% of its volume and not exceed the weight limit