Problem 1: Find the absolute extrema of function f (x, y) = 3x2 + 2y 2 − 4y on
the region bounded by the curves C1 = {y = x2 } and C2 = {y = 4}.
5
4
3
2
1
0
-3
-2
-1
0
1
2
3
(a) Regions and candidate points
(b) Graph of the function
Solution:
1. Find the critical points in the interior:
fx = 6x = 0
=⇒ x = 0, y = 1.
fy = 4y − 4 = 0
There is a unique critical point (0, 1) in the interior. Note that f (x, y) =
3x2 + 2(y − 1)2 − 2 ≥ −2 and f (0, 1) = −2. So we immediately get (0, 1)
is an absolute minimum point. We can compute the value:
f (0, 1) = −2.
2. Restrict f (x, y) to the boundary:
• Restrict f (x, y) to the curve C1 . g1 (x) = f (x, x2 ) = 2x4 − x2 on the
interval [−2, 2].
g10 (x) = 8x3 − 2x = 2x(4x2 − 1) = 0 =⇒ x = 0, ±1/2.
We get the following candidate points corresponding the critical points
in the interior of the interval: (0, 0), (1/2, 1/4), (−1/2, 1/4). There
are extra two candidates corresponding to the ends points of the interval [−2, 2]: (−2, 4), (2, 4). We compute the values at candidate
points:
f (0, 0) = g1 (0) = 0;
1
f (±1/2, 1/4) = g1 (±1/2) = − ;
8
f (±2, 4) = g1 (±2) = 28.
• Restrict f (x, y) to the curve C2 . We define g2 (x) = f (x, 4) = 3x2 +16.
g20 (x) = 6x = 0 → x = 0. So we get the following candidate points:
(0, 4), (−2, 4) and (2, 4). The latter two coming from end points of
the interval [−2, 2] and were already counted.
f (0, 4) = g2 (0) = 16.
1
3. Calculate the value of f (x, y) at candidate points and compare to find
absolute extrema. From above, we get altogether 7 candidate points:
(0, 1), (0, 0), (−1/2, 1/4), (1/2, 1/4), (0, 4), (−2, 4), (2, 4).
By the above calculation of the values of f at these candidate points, we
get that the absolute maxima is f (±2, 4) = 28. The absolute minimum is
f (0, 1) = −2.
Problem 2 (1710 = 1910 ): The problem is translated to the following optimization problem: Minimize the function
p
p
f (x, y) = 3 x2 + 4 + 2 1 + (y − x)2 + 10 − y
on the region {0 ≤ x ≤ y ≤ 10}.
12
3
10
8
2
6
4
1
2
2
4
6
8
10
2
(c) Cost optimizing configuration
4
6
8
10
12
(d) Regions and candidate points
(e) Graph of the function
1. Find the critical point in the interior:

3x
 fx = √ 3x 2 + √ 2(x−y) 2 = 0
√
=1
1
1
1
4+x
1+(x−y)
4+x2
⇒
⇒ x = √ , y = √ +√ .
2(y−x)
2
 fy = √
3(y
−
x)
=
1
−
1
=
0
2
2
3
2
1+(x−y)
We calculate the value at this critical point:
√
√
1
1
1
f ( √ , √ + √ ) = 10 + 4 2 + 3 ≈ 17.39.
2
2
3
2. Restrict f (x, y) to the boundary.
2
p
• Restrict f (x, y) to C1 = {x = 0}. g1 (y) = f (0, y) = 2 1 + y 2 −y+16
on [0, 10].
1
2y
− 1 = 0 =⇒ y = √ .
g10 (y) = p
2
3
1+y
√
So we get candidate points: (0, 1/ 3), and (0, 0), (0, 10) (two end
points). Calculate the values:
√
√
√
f (0, 1/ 3) = g1 (1/ 3) = 3 + 16 ≈ 17.73;
√
f (0, 0) = g1 (0) = 18; f (0, 10) = g1 (10) = 2 101 + 6 ≈ 26.10.
√
• Restrict
f (x, y) to C2 = {y = 10}. g2 (x) = f (x, 10) = 3 x2 + 4 +
p
2 1 + (10 − x)2 on [0, 10]. To find critical point on this interval, we
can differentiate:
g20 (x) = √
3x
2(x − 10)
= 0.
+p
2
x +4
1 + (10 − x)2
It’s not easy to find the solution to this equation. One could use
computer to get approximate minimal point:
f (1.76557) ≈ 24.59.
Also for the new corner point (10, 10), we calculate
√
f (10, 10) = g2 (10) = 3 104 + 2 ≈ 32.59.
√
• Restrict f (x, y) to C3 = {y = x}. g3 (x) = f (x, x) = 3 x2 + 4−x+12
on [0, 10].
3x
1
g30 (x) = √
− 1 = 0 =⇒ x = √ .
2
x2 + 4
√
√
So we get one new candidate point: (1/ 2, 1/ 2). Calculate
√
√
√
√
f (1/ 2, 1/ 2) = g3 (1/ 2) = 12 + 4 2 ≈ 17.66.
3. By comparison of the above values, we find the absolute minimum:
√
√
1
1
1
f √ ,√ +√
= 10 + 4 2 + 3 ≈ 17.39.
2
2
3
3