SOLUTION: A cylindrical wire frame for a wastebasket has a wire circle for the top and bottom, and SIX straight wire rods that hold the top and bottom circles together. The total amount of w

Algebra ->  Test -> SOLUTION: A cylindrical wire frame for a wastebasket has a wire circle for the top and bottom, and SIX straight wire rods that hold the top and bottom circles together. The total amount of w      Log On


   

Question 822130: A cylindrical wire frame for a wastebasket has a wire circle for the top and bottom, and SIX straight wire rods that hold the top and bottom circles together. The total amount of wire used is 6m. What radius and height will maximize the volume of the cylindrical frame?
This is an optimization problem. I'm stuck /:
So far I have v=pir^2h
2pi r + 2pi r + 6h=6
Which expresses the surface area.
I tried to solve for h and got that h= 6-4pi r over 6
At this point I really don't know what to do next . Any help is greatly appreciated.
This is a grade 12 optimization problem
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
You are definitely on the right track. This is an optimization problem, and we are trying to maximize the enclosed volume.
V+=+pi%2Ar%5E2h, and the volume will be maximized where dV/dr = 0. You have already expressed h as a function of r using the information provided.
h+=+%286-4pi%2Ar%29%2F6+=+1+-+%282%2F3%29pi%2Ar
So we make the substitution before taking the derivative:
V+=+pi%2Ar%5E2%2A%281-%28%282pi%2Ar%29%2F3%29%29
Taking the derivative and setting equal to zero you will end up with an expression like 2pi%2Ar%281-pi%2Ar%29+=+0
Since r=0 is not a solution, we are left with 1-pi%2Ar+=+0 or r+=+1%2Fpi
Now solve for h:
h+=+1+-+%282%2F3%29pi%2Fpi+=+1%2F3
The graph of the volume as a function of r is shown below. We can see the volume maximizes at r = 1%2Fpi, a little over 0.3.
graph%28300%2C300%2C-1%2C2%2C-1%2C1%2Cpi%2Ax%5E2%2A%281-%282%2F3%29%2Api%2Ax%29%29