COURSE: MATH 1500
DATE & TIME: ,
UNIVERSITY OF MANITOBA
Test 9
DURATION:
PAGE: 1 of 6
1.[9] Solution:
Let lbe the length and the width and hbe the height.
Since the volume is 8, we know that l2h= 8 and thus h=8
l2.
We want to minimize surface area which is
S= 2lw + 2lh + 2hw = 2l2+ 4lh = 2l2+32
l
Hence the function is S(l)=2l2+32
lwith domain (0,∞).
Taking the derivative yields
S0(l) = 4l−32
l2which is 0 when l3= 8 ⇒l= 2.
Since
S00
(
l
) = 4 + 64
l−3
which is positive when
l
= 2, we know we have a local
minimum. Since there is only one critical number on the interval, it’s an absolute
minimum.
Since at
l
= 2,
h
=
8
22
= 2, the dimensions which minimize surface area is when the
length, width and height are all 2m.
2.[9] Solution:
Let lbe the length and the width and hbe the height.
Since the surface area is 2400, we know that l2+ 4lh = 2400 ⇒h=2400 −l2
4l.
We want to maximize volume which is
V=l2h=1
4(2400l−l3)
Hence the function is V(l) = 1
4(2400l−l3) with domain (0,√2400).
Taking the derivative yields
V0(l) = 1
4(2400 −3l2) which is 0 when l2= 800 ⇒l=√800.
COURSE: MATH 1500
DATE & TIME: ,
UNIVERSITY OF MANITOBA
Test 9
DURATION:
PAGE: 2 of 6
Since
V00
(
l
) =
1
4
(
−
6
l
) which is negative when
l
=
√800
, we know we have a local
maximum. Since there is only one critical number on the interval, it’s an absolute
maximum.
Since at
l
=
√800
,
h
=
400
√800
and
V
=
l2h
= 400
√800
, the maximum volume is
400√800m3.
