SOLUTION: 1) 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
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Question 52991: 1) 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.
Answer
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 Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! 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.
