428 algebra De mystif ieD
- 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.