SOLUTION: I need to solve the problem in this link please https://drive.google.com/open?id=0B2SbxAb-ajtfRW00d2NCZWdFSW8

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I need to solve the problem in this link please https://drive.google.com/open?id=0B2SbxAb-ajtfRW00d2NCZWdFSW8      Log On


   



Question 1084031: I need to solve the problem in this link please
https://drive.google.com/open?id=0B2SbxAb-ajtfRW00d2NCZWdFSW8

Answer by ikleyn(52866) About Me  (Show Source):
You can put this solution on YOUR website!
.
The length of the rectangle is 2x (from -x to +x). It is "horizontal" dimension.


The height of the rectangle is f(x) = 8+-+2x%5E2. It is "vertical" dimension.


The area of the rectangle is  A(x) = 2x%2A%288-2x%5E2%29 = 16x+-+4x%5E3.


To find the maximum, take the derivative and equate it to zero:

%28dA%28x%29%29%2F%28dx%29 = 16 - 12x^2 = 0.

====>  x%5E2 = 16%2F12  ====>  x = sqrt%2816%2F12%29 = 4%2F%282%2Asqrt%283%29%29 = %282%2Asqrt%283%29%29%2F3.


You calculate maximal area A(x) by substituting x = %282%2Asqrt%283%29%29%2F3.


graph%28+330%2C+330%2C+-2.5%2C+5.5%2C+-12.5%2C+15.5%2C%0D%0A++++++++++16x+-+4x%5E3%0D%0A%29

Plot A(x) = 16x+-+4x%5E3