SOLUTION: Find the dimension and area of the rectangle with the greatest area that can be implemented in a semi-circle with radius 3 m.

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Find the dimension and area of the rectangle with the greatest area that can be implemented in a semi-circle with radius 3 m.      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 1034416: Find the dimension and area of the rectangle with the greatest area that can be implemented in a semi-circle with radius 3 m.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
I will give only an answer in which one side of the rectangle is along the diameter of the semi-circle. This way the center actually cuts the said side into two parts of equal length. Let one-half of the side have length x meters, and the height y cm. Hence, x%5E2%2By%5E2+=+3%5E2+=+9, or y+=+sqrt%289+-+x%5E2%29
The area is then given by A = 2xy, or A+=+2x%2Asqrt%289+-+x%5E2%29.
Get the derivative of A wrt x and set to 0:
sqrt%289+-+x%5E2%29+-+x%5E2%2Fsqrt%289+-+x%5E2%29+=+0
<==> 9-x%5E2+=+x%5E2 ==> 2x%5E2+=+9, or x+=+3%2Fsqrt%282%29.
If x+%3C+3%2Fsqrt%282%29, then A' > 0.
If x+%3E+3%2Fsqrt%282%29, then A' < 0.
==> there is local max at x+=+3%2Fsqrt%282%29.
Incidentally, the maximizing dimensions are x+=+3%2Fsqrt%282%29+=+y, and the maximum area is A=+2%2A%283%2Fsqrt%282%29%29%2A%28+3%2Fsqrt%282%29%29=+9+cm%5E2.