Asked by Blare
A container with square base, vertical sides, and open top is to be made from 200 ft^2 of material. Find the dimensions (height and length of base) of the container with greatest volume. If appropriate, leave your answer in radical form and enter all fractions in lowest terms.
Answers
Answered by
oobleck
as I told you six hours ago -- did you even bother to look?
If the base has sides x, and the height is h, then the area is
x^2 + 4xh = 200
so, h = (200-x^2)/(4x) = 50/x - x/4
Now the volume is
v = x^2 h = x^2 (50/x - x/4) = 50x - x^3/4
Now just find where dv/dx = 0 for maximum volume.
Post your work if you get stuck
If the base has sides x, and the height is h, then the area is
x^2 + 4xh = 200
so, h = (200-x^2)/(4x) = 50/x - x/4
Now the volume is
v = x^2 h = x^2 (50/x - x/4) = 50x - x^3/4
Now just find where dv/dx = 0 for maximum volume.
Post your work if you get stuck
Answered by
Blare
I did check. It just closed the window and I didn’t remember exactly what you said and I couldn’t find the post so I reposted. Sorry.
As for my work, I tried solving for dv/dx and I get 50 - 3/(4(x^1/4)) and when I plug 0 in for that I get 81/1600000000. But how would I use this in order to find the specifics involving the length for the base of the box and the height? You show the formula for the height but is this number x or?
As for my work, I tried solving for dv/dx and I get 50 - 3/(4(x^1/4)) and when I plug 0 in for that I get 81/1600000000. But how would I use this in order to find the specifics involving the length for the base of the box and the height? You show the formula for the height but is this number x or?
Answered by
oobleck
Really? You need to read more carefully ...
that's not x^(3/4) It is (x^3)/4
dv/dx = 50 - 3/4 x^2
that's not x^(3/4) It is (x^3)/4
dv/dx = 50 - 3/4 x^2
Answered by
Blare
Okay that makes much more sense. Sorry for misreading. Thanks for the help. I figured it out now.