Answer to Question #260275 in Geometry for niel

Question

a pottery manufacturer has an order to manufacture 5000 hanging vases to hold ½ pt. of water when full. The vases are designed so as to fit into the corner of a room. The faces of each vase are triangular in shape and intersect to form a pointed bottom. The area of the polygon cut out of the plane of the base by the lateral faces is 3sq.in. The height of the vase is 8 in. Compute the weight of pottery required if pottery weighs 130lb. per cu.ft.

Answer 1

$$V_ {out}= \dfrac{1}{3}Sh=\dfrac{1}{3}(3in^2)(8in)=8in^3$$

Vase holds ½ pt. of water when full.

$$V_{in}=1/2\ pt=14.4375in^3$$

Since $V_ {out}=8in^3<14.4375in^3=V_{in},$ then there is no solution.

Such vase is impossible.

Let us redefine the task:

a pottery manufacturer has an order to manufacture 5000 hanging vases to hold 1/4 pt. of water when full. The vases are designed so as to fit into the corner of a room. The faces of each vase are triangular in shape and intersect to form a pointed bottom. The area of the polygon cut out of the plane of the base by the lateral faces is 3sq.in. The height of the vase is 8 in. Compute the weight of pottery required if pottery weighs 130lb. per cu.ft.

$$V_ {out}= \dfrac{1}{3}Sh=\dfrac{1}{3}(3in^2)(8in)=8in^3$$

Vase holds 1/4 pt. of water when full.

$$V_{in}=1/4\ pt=7.21875in^3$$

Then

$$V=V_{out}-V_{in}$$

$$=8in^3-7.21875in^3=0.78125in^3$$

$$1 lbs/ft ^3=453.59237 g/(1728in^3)$$

$$m_{vase}=130(\dfrac{453.59237 g}{1728in^3})(0.78125in^3)=26.6597g$$

$$M=5000(26.6597g)=133298.540g$$

The total weight of the 5000 vases is 133 kg 299 g.

In this case we obtain the answer.