Optimization: What is the maximum capacity of a box

[wind]

Hi i tryed to solve this problem but my anwsers don't make sense...could someone help me with this? Thanks

An open metal box for removing ashes from a fireplace is to be constructed form a rectangular pice of sheet metal that is 1m by 1.5m. Suqares are to be cut from each corner of the sheet metal, the sides folded upward to form the box, and then the seams welded. What is the maximum capacity of a box that is constructed in this way?

v=Lwh

L= 1-2x
w= 1.5-2x
h= x

v=(1-2x)(1.5-2x)(x)
v=(1.5-2x-3x+4x)(x)
v=1.5x-x²

v'=1.5-2x
0=1.5-2x
-1.5=-2x
0.75=x

L= 1-2(0.75)=-0.5
w= 1.5-2(0.75)=0
h= 0.75
v=0

:?

 

[tkhunny]

Don't worry. Lots of calculus students trip over algebra. You should not have continued when your volume had only x^2. You multiplied THREE things. You should get x^3. Try that multiplication again. ALWAYS be in the lookout for things that don't look right.

For example, if you get the anser x = 0.77 m, will you believe it? How about x = 1.7 m? Why or why not?

ALWAYS be on the lookout.

 

[wind]

Thanks tkhunny

v=(1-2x)(1.5-2x)(x)
v=(1.5-2x-3x+4x²)(x)
v=(1.5-5x+4x²)(x)
v= 1.5x-5x²+4x³


v'=1.5-10x+12x²
0=12x²-10x+1.5

(10+root28)/24
(10-root28)/24

x=0.63715
x=0.19619

x=0.63715 inadmissable

L= 1-2(0.19619) =0.60762
w= 1.5-2(0.19619)=1.10762
h= 0.19619

v=(0.60762)(1.10762)(0.19619)
v=0.132038