A container with a square base, vertical sides and an open top is to be made from 1200ft squared of material. find the dimensions of the container with the greatest volume.

Question

anonymous · Answer

how do i begin

amistre64 · Answer

Do you know if all 1200 ft^2 of material is to be used?  or is it one of those"cut out the corners" kind of problems?

amistre64 · Answer

the container, if all material is to be used would have a surface area of:

4(side area) + (base area) = 1200

anonymous · Answer

you just need the sum of the areas of the surfaces to add up to 1200sqft.  so (area of base)+4×(area of one side) = 1200 sqft

amistre64 · Answer

4xy + x^2 = 1200 , does that sound right?

amistre64 · Answer

Volume = x^2 y

amistre64 · Answer

solve for one variable in terms of the other, plug it into the volume equation, and derive :)

amistre64 · Answer

y = (1200 -x^2)/4x

amistre64 · Answer

and dont let the "hero" title fool ya.... I am after all an idiot in disguise :)

amistre64 · Answer

V = (x^2/4x)(1200-x^2)

amistre64 · Answer

V = (1/4) (1200-x^3)  if I did it right...

amistre64 · Answer

*1200x - x^3

amistre64 · Answer

dV = (1/4)(1200 - 3x^2)   maybe  make that equal to 0 to get the critical numbers

anonymous · Answer

Oh, I misread the problem.. yes, put y in terms of x, y = (1200 -x^2)/4x, then substitute into volume V = x^2 y = x^2 (1200 -x^2)/4x = x(1200-x^2)/4 = x(300-(x^2)/4), then find the maximum volume

amistre64 · Answer

1200 = 3x^2

amistre64 · Answer

x = sqrt400)

anonymous · Answer

If you've taken calculus you can do that my finding the zeros of the derivative, otherwise I think you'll have to test sample values and just find the x that yields the greatest V

amistre64 · Answer

x=20 i think

plug that back in to your "y" equation to ge the value for y

amistre64 · Answer

y = (1200 -20^2)/4(20)

y = (1200 - 400)/ 80

y = 800/80 = 80/8 = 10/1

y = 10 and x = 20  if i did it all correctly

amistre64 · Answer

determine if 20x20x10  is greater than 10x10x20 and youve got your answer

amistre64 · Answer

or.... ignore that last comment...its probably my stupidity talking :)

amistre64 · Answer

20x20 for the base, and 10 high fits all the requirements

amistre64 · Answer

but, do you see how we got it?

anonymous · Answer

yeah

anonymous · Answer

square roots :)