to find maximum find v'=0
12x² -128x + 256 = 0
4(3x² -32x + 64) = 0
Need help from here
You are correct so far . Now solve for x . Then you will get 2 values of x so you will need to verify which of them would give a max volume .
Use second order : d^2y/dx^2=6x-32
Whichever which gives a negative value is the one .
Then substitute this value of x back into the original equation
V=4x³ - 64x² + 256x
July 30th 2009, 06:37 AM
You are OK so far I think. The quadratic is probably going to factorise and x=8 will be a root, since the volume of the box is clearly 0 then and will then start increasing again.
In fact (3x^-2-32x+64) = (x-8)(3x-8).
So now you can easily calculate the volume.
July 30th 2009, 07:00 AM
Roots are 8 and 8/3
back into the equation 8 gives a volume of 0, so max volum is where x=8/3