Question 33660
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...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, -2, 3, -2, 200, 4*(x^3)-28*(x^2)+48*x)}}}
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The volume of a cylinder (think about the volume of a can) is given by V = pr2h where r is the radius of the cylinder and h is the height of the cylinder. Suppose the volume of the can is 100 cubic centimeters.
a)Write h as a function of r.
Answer
100=PI*R^2*H
H=100/(PI*R^2)
b)What is the measurement of the height if the radius of the cylinder is 2 centimeters?
Answer
H=100/(PI*2^2)=25/PI=7.96 CM.

Show work in this space

c)Graph this function.
Show graph here
{{{graph( 500, 500, -2, 10, -2, 10, 100/(3.14*(x^2)))}}}