An Example 2D Optimization Problem
Let D be the closed region defined by x ≥ 0, y ≥ 0, y ≤ x, x2 + y 2 ≤ 4. D is
pictured below.
Figure 1: The region D.
Let f (x, y) = y 2 − y + x2 − 3x. What are the maximum and minimum values
of f on the region D, and where do they occur?
Solution:
The process will always look something like this:
1. Figure out where the boundary curves meet, i.e., the “corners.” Place
these in a list of “candidate” extremal points.
2. Find the critical points of f by computing partial derivatives and setting
them equal to zero (or places where the derivative does not exist). Add
these to the list of candidate points.
3. Find the critical points of f , considered as a single-variable function, on
each of the boundary curves. This entails parameterizing each curve as
some (g(t), h(t)), t1 ≤ t ≤ t2 ; computing F (t) = f (g(t), h(t)); and finding
its critical points in the interval [t1 , t2 ] by setting dF/dt = 0, solving for t,
and then plugging these values of t into the function F . Add these points
to the list of candidate points.
4. Throw out any candidate points which are not in the region under consideration.
5. For each candidate point (a, b), compute f (a, b). Whichever point(s)
give(s) the largest value for f is/are the global maximum/maxima; the
point(s) achieving the smallest value for f is/are the global maximum/maxima.
1
First, we note√the boundary curves. One is y = 0; another is y = x, and
the third is y = 4 − x2 , since that is what one obtains by solving for y in
x2 + y 2 = 4. The first two
√ curves clearly meet at (0, 0). The first and third
meet at the point where 4 − x2 = 0, i.e.,
0). The second and
√ the point (2,
2 , i.e., x2 = 4 − x2 , so 2x2 = 4,
third curves meet at the point where
x
=
4
−
x
√ √
x2 = 2, and we
the point ( 2, 2). So far, our candidate point list is
√
√ obtain
{(0, 0), (2, 0), ( 2, 2)}.
Next, we find the critical points of f . We have fx (x, y) = 2x − 3 and
fy (x, y) = 2y − 1. Therefore, if we solve 2x − 3 = 0 and 2y − 1 = 0, we obtain
the point (3/2, 1/2). Adding this to the candidate list gives
√ √
{(0, 0), (2, 0), ( 2, 2), (3/2, 1/2)}.
Now, we parameterize the three curves that form the boundary of D: y = 0
can be parameterized
√ (t, 0), 0 ≤ t ≤ 2; y = x can be parameterized (t, t),
√
0 ≤ t ≤ 2; and √ 4 − x2 can be parameterized (2 cos θ, 2 sin θ), 0 ≤ θ ≤ π/4.
(We could use (t, 4 − t2 ), but I went with the trigonometric parameterization
for a little variation.) Along the first curve, F1 (t) = f (t, 0) = t2 − 3t. This has
critical points when 2t − 3 = 0, i.e., t = 3/2. In that case, the point is (3/2, 0).
Along the second curve, F2 (t) = f (t, t) = 2t2 − 4t. This has critical points when
4t − 4 = 0, i.e., t = 1. In that case, the point is (1, 1). Along the third curve,
F3 (θ) = 4 sin2 θ − 2 sin θ + 4 cos2 θ − 6 cos θ. We have
dF3
= 8 sin θ cos θ − 2 cos θ − 8 sin θ cos θ + 6 sin θ = 6 sin θ − 2 cos θ.
dθ
The derivative is zero when 3 sin θ = cos θ, which
p unfortunately cannot be solved
explicitly
for
θ.
However,
since
cos
θ
=
(±)
1 − sin2 θ, we may write 3s =
√
2
1 − s , where
s = sin θ. Squaring both sides yields 9s2 = 1 − s2 , i.e., 10s2 = 1,
√
so s = 1/ 10. Since the point we are concerned with is (2 cos θ, 2 sin θ), the
critical point is
 s

2
6
2
2
2 1 − √1
,√ = √ ,√
.
10
10
10
10
So, if we add all of these critical points to the growing list, we obtain
√ √
√
√
{(0, 0), (2, 0), ( 2, 2), (3/2, 0), (1/2, 0), (1, 1), (6/ 10, 2/ 10)}.
It is easy to verify that all of these points are in D. Therefore, we need only
compare the value of f at these points.
f (0, 0)
√f (2,
√0)
f ( 2, 2)
f (3/2, 1/2)
f (3/2, 0)
f√
(1, 1)
√
f (6/ 10, 2/ 10)
2
=
=
=
=
=
=
=
0
−2 √
4−4 2
−5/2
−9/4
−2 √
4 − 2 10.
These points are pictured in Figure 2 on the graph of f , along with the
region D.
Figure 2: The region D and the graph of f on the region D. The three boundary
curves are drawn on both D and the graph in purple. “Corners” are labeled in
red. The yellow points are all of the candidate extrema.
Note that all of the values are
case of −2, −5/2,
√ ≤ 0. This is clear in the √
and −9/4.
To see that 4 − 4 2 < 0, simply note that 2 > 1. Finally,
√
4 − 2 10 < 4 − 2 · 3 = −2 < 0, where the first inequality follows from the fact
that 32 = 9 < 10. Hence, 0 is the largest value, and our function achieves
its maximum value of 0 at the point (0, 0). The following ordering is easy
to check:
5
9
− < − < −2 < 0,
2
4
and so certainly
−9/4,
−2,
and
0
are
not
global minima. We need only compare
√
√
−5/2, 4 − 4 2, and 4 − 2 10. First, note that
√
√
√
4 − 4 2 = 4 − 2 8 > 4 − 2 10,
√
so 4 − 4 2 cannot be our minimum. Lastly,
√ √
5 13
− − 4 − 2 10 = 2 10 −
< 0,
2
2
√ 2
2
because
√ (13/2) = 169/4 = 42 + 1/4 > 40 = (2 10) . Therefore, −5/2 <
4 − 2 10, so the minimum of the function f on the domain D is −5/2
and is achieved at the point (3/2, 1/2).
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