SOLUTION: Find the minimum dimensions of an open-top rectangular box that is twice as long as it is wide and can hold 12 cubic yards of material.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find the minimum dimensions of an open-top rectangular box that is twice as long as it is wide and can hold 12 cubic yards of material.      Log On


   

Question 829230: Find the minimum dimensions of an open-top rectangular box that is twice as long as it is wide and can hold 12 cubic yards of material.
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
x=length
y=width
z=height

x=2y; base area is xy=%282y%29y=2y%5E2;
volume, xyz=12
z2y%5E2=12
zy%5E2=6
z=%281%2Fy%5E2%296 or z=6y%5E%28-2%29

This still has two variables. Both unknown but one of them depends on the other. This has no minimum set of dimensions. Choose any positive z and it determines the needed positive y; or choose any positive y and this determines z.

Even if you try finding derivative and looking for some extreme point, none is!

z as a function of y:
graph%28300%2C300%2C-1%2C10%2C-1%2C10%2C6x%5E%28-2%29%29