SOLUTION: An open-top box is to be constructed from a 6 foot by 8 foot rectangular cardboard by cutting out equal squares at each corner and the folding up the flaps. Let x denote the length

Question 64192: An open-top box is to be constructed from a 6 foot by 8 foot rectangular cardboard by cutting out equal squares at each corner and the folding up the flaps. Let x denote the length of each side of the square to be cut out.
a) Find the function V that represents the volume of the box in terms of x.

b) Graph this function and show the graph over the valid range of the variable x..
Show Graph here

c) Using the graph, what is the value of x that will produce the maximum volume?

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!

Volume/44472: An open-box top is to be constructed
from a 6 by 8 foot rectangular cardboard by cutting
out equal squares at each corner and then folding up
the flaps. Let x denote the length of each side of the
square to be cut out.
A) Find the function V that represents the volume of
the box in terms of x.
B) Graph this function and show the graph over the
valid range of the variable x.
C) Using the graph, what is the value of x that will
produce the maximum volume?
I have been struggling with this problem all week if
it wasn't for this problem my assignment would be
done. Please help if you can thanks!!!
1 solutions
Answer 29381 by venugopalramana(2734) About Me on
2006-07-01 02:06:41 (Show Source):
You can put this solution on YOUR website!
An open-box top is to be constructed from a 6 by 8
foot rectangular cardboard by cutting out equal
squares at each corner and then folding up the flaps.
Let x denote the length of each side of the square to
be cut out.
A) Find the function V that represents the volume of
the box in terms of x.
B) Graph this function and show the graph over the
valid range of the variable x.
C) Using the graph, what is the value of x that will
produce the maximum volume?
WHEN WE CUT X LONG PIECES ON ALL 4 SIDES THE CARD
BOARD WILL GET REDUCED BY
X+X=2X...ALONG LENGTH AND...X+X=2X.....ALONG WIDTH
SO OPEN BOX LENGTH = 8-2X AND WIDTH = 6-2X..AND HEIGHT
=X ...SO VOLUME V IS GIVEN BY LEMGTH*WIDTH*HEIGHT
V=(8-2X)(6-2X)X...DOMAIN OF V IS GIVEN BY THE FACT
THAT LENGTH OR WIDTH CAN NOT BE NEGATIVE...CRITICAL
VALUE BEING WIDTH WE GET ....
8-2X>0...AND 6-2X>0...SO X <3
RANGE.....MAXIMUM VALUE........
V=X(8-2X)(6-2X)=X{48-16X-12X+4X^2)=4X^3-28X^2+48X...
--------------------------------------------------------------------
OMIT THIS PORTION IF YOU DO NOT KNOW CALCULUS
------------------------------------------------------------
IF YOU KNOW CALCULUS
DV/DX=12X^2-56X+48=0..OR...3X^2-14X+12=0....
X=(14+SQRT.(52))/6...OR......(7+SQRT.(13))/3...OR....(7-SQRT.13)/3
X=3.54..OR...1.13.
D2V/DX2=6X-14=- VE AT X=1.13...SO MAXIMUM VOLUME IS
OBTAINED AT X=1.13'
-----------------------------------------------------------------------------
YOU CAN SEE THAT MAXIMUM VOLUME OCCURS AT X=1.13 BY
PLOTTING THE GRAPH.

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