SOLUTION: For the anniversary celebration, there will be helium-filled balloons for the children. They have purchased spherical balloons that have a 9 in. diameter. When filled with heli

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: For the anniversary celebration, there will be helium-filled balloons for the children. They have purchased spherical balloons that have a 9 in. diameter. When filled with heli      Log On


   

Question 1182311: For the anniversary celebration, there will be helium-filled balloons for the children.
They have purchased spherical balloons that have a 9 in. diameter.
When filled with helium, the balloons should continue to fly for most of the day.
The committee estimates that they will need 1500 balloons.
Part 1:
They plan to purchase the cylinders of helium and fill the balloons themselves.
They have two choices of cylinders:
a medium cylinder which contains 80 cu. ft. of helium
a large cylinder which contains 100 cu. ft. of helium
The committee does not want to purchase too much extra helium, but they need to have enough to fill all 1500 balloons.
How many balloons can be filled using a medium cylinder?
How many balloons can be filled using a large cylinder?
How much helium is needed to fill all the balloons?
Advise the committee of which cylinder(s) to purchase and how many of each they should order. (Ignore cost.)

Outline your plan:

Complete your calculations:

Advise the committee:
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Find the volume of a balloon with 9 in diameter or 4.5 inch radius. That is (4/3)*π*4.5^3
=381.70 in^3/balloon
this is 0.22089 cu. ft. of helium
80/0.22089=362.17 balloons per 80 ft^3 (362)
100/0.22089=452.71 balloons per 100 ft^3 (452)
An way to decrease waste would be 3 of the medium (1086) and 1 of the large (452), which would fill 1538 balloons.


Answer by ikleyn(52873) About Me  (Show Source):
You can put this solution on YOUR website!
.
For the anniversary celebration, there will be helium-filled balloons for the children.
They have purchased spherical balloons that have a 9 in. diameter.
When filled with helium, the balloons should continue to fly for most of the day.
The committee estimates that they will need 1 500 balloons.
Part 1:
They plan to purchase the cylinders of helium and fill the balloons themselves.
They have two choices of cylinders:
a medium cylinder which contains 80 cu. ft. of helium
a large cylinder which contains 100 cu. ft. of helium
The committee does not want to purchase too much extra helium, but they need to have enough to fill all 1500 balloons.
How many balloons can be filled using a medium cylinder?
How many balloons can be filled using a large cylinder?
How much helium is needed to fill all the balloons?
Advise the committee of which cylinder(s) to purchase and how many of each they should order. (Ignore cost.)
~~~~~~~~~~~~~~~~~~~ 

One balloon volume = %284%2F3%29%2Api%2Ar%5E3 = %284%2F3%29%2A3.14159%2A4.5%5E3 = 381.7032 in^3 = 381.7032%2F12%5E3 = 0.220893047 ft^3.


1500 ballons volume = 1500*0.220893047 ft^3 = 331.3395703  ft^3  - total necessary volume of helium for 1500 balloons.



The further part of the problem is posed uncertainly; but we can interpret it as the requirement to have the total volume 
of the helium above 331.33957 ft^3, but minimize the extra helium.


More formally, we should have


    80x + 100y >= 331.33957  ft^3  of helium

      x >= 0,  y >= 0,  where  x  and  y  are integer numbers;

    80x + 100y - 331.33957  --->  minimum.


The simplest way to solve this minimization problem is to make a table of reasonable values of x and y 
(the numbers of cylindrical containers of both types) and determine the necessary minimum from the table.



       T     A     B     L     E


    x     y     80x + 100y    the extra helium volume
                              80x + 100y - 331.33957

    5	  0	  400	          68.66043
    4	  1	  420	          88.66043
    3	  1	  340	           8.66043
    2	  2	  360	          28.66043
    1	  3	  380	          48.66043
    0	  4	  400	          68.66043


The numbers in the first two column are chosen in this way to cover all reasonable options
and do not consider (cut off) all UNREASONABLE versions.


The calculations show that all these options provide the necessary volume of the helium
and the minimum of extra helium is at  x = 3  (three 80 ft^3 helium containers) and y = 1 (one 100 ft^3 helium container).


ANSWER.  Among all possible combinations of the helium containers, that provide

         the necessary minimum helium volume and minimize extra volume of helium, 

         the solution is  three 80 ft^3 helium containers and one 100 ft^3 helium container.

Solved.

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It is how I would solve it as a Math problem; how I would present the arguments
and how I'd made the final presentation for the higher authority to consider and to approve the project.