Algebra Demystified 2nd Ed

428 algebra De mystif ieD

5. Because the can’s diameter is 6, the radius is 3. Let x represent the
   increase in the radius of the can. The radius of the new can is 3 + x.
   The volume of the current can is V = or^2 h = o(3)^2 5 = 45o. To increase
   the volume by 50% means to add half of 45π to itself; the new volume
   would be 45 1
   2

45 90

2

45

2

135

2

oo+=oo+=o. The volume formula for the

new can becomes^135

2

o=+o() 35 x (^2).
135
2
35
1
5
135
2
3
27
2
3
27
2
3
2
2
2
o o
o
o
=+
=+
=+
=+
⋅
()
()
()
(
x
x
x
xxx
xx
xx
)( )
()
3
27
2
96
227
2
29 6
27 18 1
2
2

• =+ +
  
  
  
  
  =++
  =+ 222
  27 21218
  02 12 9
  12 12 42
  2
  2
  2
  2
  xx
  xx
  xx
  x


• 
  

  =++
  =+−
  =−± −−()(^99
  22
  12 144 72
  4
  12 216
  4
  12 66
  4
  12 66
  4
  2
  )
  ()
  =−± +
  =−± =−± ⋅ =−±
  =^22636
  4
  636
  2
  ()−± =−± ≈0 674.
  (The other solution is negative.)
  The manufacturer should increase the can’s radius by about 0.674 inches.
  Because the diameter is twice the radius, the manufacturer should
  increase the can’s diameter by about 2(0.674) = 1.348 inches.