Question 52991
An open-top box is to be constructed from a 6 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.
c)Using the graph, what is the value of x that will
produce the maximum volume?
Thank you for your help. !!!!! ;')
1 solutions
Answer 19558 by venugopalramana(1619) About Me on
2006-04-10 12:13:41 (Show Source):
I AM ABLE TO FIND ONE SUCH QUESTION I ANSWERED
EARLIER.SEE THAT AND BY THE SIDE CORRESPONDING ANSWER
TO YOUR QUESTION
Volume/30504: an open box is to be constructed from a
piece of cardboard 15 inches by 25 inches by cutting
squares of length x from each corner and folding up
the sides. Express the volume of the box as a function
of x. what is the domain v?
1 solutions .........IN YOUR CASE THE DIMENSIONS ARE
6' AND 8'


Answer 17192 by venugopalramana(1167) About Me on
2006-03-17 06:08:55 (Show Source):
an open box is to be constructed from a piece of
cardboard 15 inches by 25 inches by cutting squares of
length x from each corner and folding up the sides.
Express the volume of the box as a function of x. what
is the domain v?
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 = 25-2X ..(IN YOUR CASE 8-2X) AND
WIDTH = 15-2X..(IN YOUR CASE 6-2X)..AND HEIGHT =X
...SO VOLUME V IS GIVEN BY LEMGTH*WIDTH*HEIGHT
V=(25-2X)(15-2X)X...(IN YOUR CASE
(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 ....
15-2X>0...OR....15>2X...OR....7.5>X....OR X<7.5...(IN
YOUR CASE 8-2X>0...AND 6-2X>0...SO X <3)
RANGE.....MAXIMUM VALUE....IN YOUR CASE....
V=X(8-2X)(6-2X)=X{48-16X-12X+4X^2)=4X^3-28X^2+48X...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 IT BY PLOTTING THE GRAPH.
{{{graph( 500, 500, 0, 3, 0, 200, 4*(x^3)-28*(x^2)+48*x)}}}