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 o

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Question 26217: 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.
Show Graph here



c) Using the graph, what is the value of x that will produce the maximum volume?
Answer
Found 2 solutions by venugopalramana, danirivera5:
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
An open-top box is to be connected 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. Find the function V that represents the volume of the box in terms of x.
This is what I got:
6x+8x+x=14x²
x=14
LENGTH OF BOARD=6 FT....WIDTH OF BOARD =8 FT.
IF X LONG SQUARE PIECES ARE CUT AT 4 CORNERS WE SHALL HAVE THE BOX DIMENSIONS AS
HEIGHT =X.....LENGTH = 6-2X.....WIDTH =8-2X
VOLUME = V = L*W*H =X(6-2X)(8-2X)=X(48-28X+4X^2)=48X-28X^2+4X^3
THE GRAPH WILL LOOK LIKE THIS
+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C48%2Ax-28%2Ax%5E2%2B4%2Ax%5E3%29+
FROM THE GRAPH YOU CAN SEE V HAS A PEAK AT X=ABOUT 3.5 FEET

Answer by danirivera5(2) About Me  (Show Source):
You can put this solution on YOUR website!
4/10/06
..(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'