A reservoir is in the form of frustum of a cone, 68 ft. across the top, 35 ft. across the bottom, and 18 ft. deep. Find the cost

Renz Kolin Gabales Cañete's picture
fitzmerl duron's picture

To answer the question above, kailangan muna natin ng figure....It looks like this:

reservoir.png

Assuming that your reservoir is a closed tank, your reservoir looks ugly because it will be lined/covered with tiles...hehehehe...

Your reservoir is shaped in a frustrum of a cone and it is lined with tiles, we need to get the surface area of the frustrum of a cone.The surface area of the frustrum of a cone is...

$$Surface \space area = \pi (r_1^2 + r_2^2 + r_1r_2\sqrt{(r_1^2 - r_2^2)^2 + h^2})$$

where $r_1$ is radius of upper base of the frustrum of a cone, $r_2$ is radius of lower base of the frustrum of a cone, and $h$ is the height of the frustrum of a cone.

With that in mind, the surface area of the frustrum of a cone would be...

$$Surface \space area = \pi (r_1^2 + r_2^2 + r_1r_2\sqrt{(r_1^2 - r_2^2)^2 + h^2})$$ $$Surface \space area = \pi ((17.5 \space ft)^2 + (34 \space ft)^2 + (17.5 \space ft)(34 \space ft)\sqrt{((17.5 \space ft)^2 - (34 \space ft)^2)^2 + (18 \space feet)^2})$$ $$Surface \space area = 50237.51 ft^2$$

The cost of lining the reservoir with tiles then is $$50237.51 ft^2 \left( \right) = \color{green}{\$ 50237.51}$$

and that is one expensive reservoir....

Alternate solutions are highly encouraged...

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