Question

Two equal rectangular sand boxes having one side in common and a square floor each, are made up of 96m^2 of material. What are the dimensions of each box if the volume is to be maximized?

Answer 1

my solution:
Area=8x^2+8xy
y=(96-8x^2)/8x

V=(4x^2)(y)
=(4x^2)((96-8x^2)/8x)
V(x)=(96x-8x^3)/2

V'(x)=(96-24x^2)/2
x=2---> it's wrong

y=(96-8(2)^2)/8(2)
=4


the correct answer is 4 x 4 x 16/7

Answer 2

@hero @thomaster @hartnn

Answer 3

Area=8x^2+8xy
how ?
a box has 6 faces, so 2 box will have 12 faces, not 16

Answer 4

A=4x^2+8xy

Answer 5

hmm.. you're right.. maybe i got confused with my drawing :/
ok,, let me solve it again..thanks!

Answer 6

and how are you calculating the volume ?

Answer 7

sorry I respond TOO late, i'm doing this question now..
uhm.. i just wanted to ask some questions,
like for the area, don't you think it should be A= A=4x^2+7xy
because the problem says "Two equal rectangular sand boxes having ONE SIDE IN COMMON"
and for the volume, I really messed up with my first solution.. now I realized that the volume should be (2x^2)y, am i right??

Answer 8

I already got the right answer (^_^)!!
since the problem says "Two equal rectangular sand boxes having ONE SIDE IN COMMON and A SQUARE FLOOR each", i finally understand that the formula for the area must be A=2x^2+7xy..
I used the formula for the area as my constraint..
then i substitute it to the volume formula which is V=(2x^2)y
then I got the right answer which is 4 x 4 x 16/7

(^_^)