MAT235, 20082009
Problem Set 4 Solutions
1.
There is nothing more annoying than studying math, so Peter is trying to minimize his study time for
passing MAT235. The overall mark will be composed by 15% from the problem sets, 45% from the term
tests and 40% from the final exam. Peter expects his mark
P
for the problem sets to be proportional to the
time
x
he spends on them (let’s say,
P
=
x
), his mark for the term tests to be
T
=
2
x
3
+
y
, where
y
is the time
he spends studying specifically for the term tests and his mark
F
for the final to be
F
=
x
2
+
y
2
+
z
, where
z
is the time he studies specifically for the final. What is the minimum amount of time for achieving the pass
mark of 50 points, and how should he break down this time?
(The above model is not realistic, because in reality the marks
P, T
and
F
cannot go higher than
100
each. Comment on how this would influence your answer.)
Solution:
Clearly, the least amount of time will go with the lowest pass score of 50, therefore we need to
minimize the function:
f
(
x, y, z
) =
x
+
y
+
z
subject to the constraint:
g
(
x, y, z
) = 0
.
15
x
+ 0
.
45
2
x
3
+
y
+ 0
.
40
x
2
+
y
2
+
z
= 50
.
If we try to apply the method of Lagrange multipliers, we will see that
∇
f
= (1
,
1
,
1)
and
∇
g
= (0
.
65
,
0
.
65
,
0
.
4)
,
so
∇
f
and
∇
g
are never proportional. (Geometrically: Both the level curves of
f
and the constraint
g
= 50
are planes, which however, are not parallel.) Therefore, there are no “critical” problems for the problem
with constraints. However, the problem has a “boundary” since all
x, y, z
are
≥
0
, and the minimum should
appear at the boundary.
We write:
f
(
x, y, z
) =
x
+
y
+
z
=
1
0
.
65
(0
.
65
x
+ 0
.
65
y
+ 0
.
40
z
) +
0
.
25
0
.
65
z
=
1
0
.
65
g
(
x, y, z
) +
0
.
25
0
.
65
z.
Since
g
(
x, y, z
) = 50
and
z
≥
0
, this is
≥
1
0
.
65
·
50 = 76
.
9
, and the minimum value of
76
.
9
is achieved when
z
= 0
. Therefore, Peter should not study at all for the final exam, and should break down a time of
76
.
9
between problem sets and term tests, in any way he wants.
The restriction that the marks cannot exceed 100 would not influence the answer, since with
76
.
9
time
units of work none of the marks
P, T
or
F
can go above
100
, no matter how Peter breaks up this time.
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 Fall '08
 Recio
 Multivariable Calculus, Sets, lagrange multipliers

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