SOLUTION: 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 e
Algebra ->
Expressions-with-variables
-> SOLUTION: 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 e
Log On
Question 33137This question is from textbook college algebra
: 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. !!!!! ;') This question is from textbook college algebra
You can put this solution on YOUR website! The dimension of the rectangular cardboard is 6 ft X 8 ft
Let x-ft be each side of the square cut out from each of the four corners.
The length of the open-top box would be (8-2x) ft (Note: two corners);
The width of the open-top box would be (6-2x) ft.
The height of the open-top box would be x ft. (if you draw a diagram, it will neatly explain the resulting dimension of the open-top box);
The voume of a box = Length X Width X Height;
If V(x) represents the volume function of the box:
V(x) = x(8-2x)(6-2x);
...............
As far as graphing goes, this is a polynomial of degree 3. This is also bound by the restriction (domain of the function) x >= 0 and x <= 3;
.....................
You need a graphing calculator to find the value of x that will produce the maximum value. You will need to use "Trace and Zoom" functionality of the graphing calculator.
You can put this solution on YOUR website! 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.
(
