Question #260275

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.


1
Expert's answer
2021-11-04T15:55:11-0400
Vout=13Sh=13(3in2)(8in)=8in3V_ {out}= \dfrac{1}{3}Sh=\dfrac{1}{3}(3in^2)(8in)=8in^3

Vase holds ½ pt. of water when full.

Vin=1/2 pt=14.4375in3V_{in}=1/2\ pt=14.4375in^3

Since Vout=8in3<14.4375in3=Vin,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.


Vout=13Sh=13(3in2)(8in)=8in3V_ {out}= \dfrac{1}{3}Sh=\dfrac{1}{3}(3in^2)(8in)=8in^3

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


Vin=1/4 pt=7.21875in3V_{in}=1/4\ pt=7.21875in^3

Then


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

=8in37.21875in3=0.78125in3=8in^3-7.21875in^3=0.78125in^3

1lbs/ft3=453.59237g/(1728in3)1 lbs/ft ^3=453.59237 g/(1728in^3)

mvase=130(453.59237g1728in3)(0.78125in3)=26.6597gm_{vase}=130(\dfrac{453.59237 g}{1728in^3})(0.78125in^3)=26.6597g

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

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

In this case we obtain the answer.


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