Question 59623: I submitted this problem earlier today but made a mistake. The correct problem is: Find the dimensions of a box whose volume is 196 cu. in., whose surface area is 280 sq. in., and whose LENGTH is twice its width.
I got 2 equations:
196=2W^2H (W is width and H is height)
280=4W^2+6WH
I solved the first equation for H and substituted it in the 2nd equation, ending up with:
4W^3-280W+588=0
I can't figure out where to go from here.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! I submitted this problem earlier today but made a mistake. The correct problem is: Find the dimensions of a box whose volume is 196 cu. in., whose surface area is 280 sq. in., and whose LENGTH is twice its width.
L=2W
V= LWH= 2W^2H=196
S.A= 2[LW+WH+HL]=280
2W^2+WH+2WH=140
2W^2+3WH=140
I got 2 equations:
196=2W^2H (W is width and H is height)..OK
W^2H=98...........H=98/W^2
280=4W^2+6WH...OK
2W^2+3WH=140
2W^2+3W*98/W^2=140
2W^3+294=140W
2W^3-140W+294=0
I solved the first equation for H and substituted it in the 2nd equation, ending up with:
4W^3-280W+588=0....OK
W^3-70W+147=0
TAKING FACTORS OF 147 ...WE HAVE 147=1*3*7*7
HENCE CHECKING F(1),F(3),F(7)
WE FIND F(7) = 7^3-70*7+147=0
HENCE W=7
L=2*7=14
H= 98/(7*7)=2
CHECK....V=14*7*2=196...OK
S.A. = 2[14*7+14*2+7*2]=2(98+28+14)=280...OK
I can't figure out where to go from here.
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