Math 280, Intermediate Calculus
Sec. 14.7 Maximum and Minimum Values
name:
date:
23. Use a graph and/or level curves to estimate the local maximum and minimum values and saddle point(s)
of the function. Then use calculus to find these values precisely.
0 ≤ x ≤ 2π,
f (x, y) = sin x + sin y + sin (x + y) ,
2Π
0 ≤ y ≤ 2π
-2
1
-1
5Π
3
0
4Π
3
2
-1
1
0
-1
-2
-1
2Π
y
z
5Π
0
Π
3
4Π
3
Π
0
Π
3
3
2
3
Π
x
3
0
Π
Π
4Π
1
3
2Π
2Π
1
2Π
y
3
3
5Π
3
2Π
1
0
0
-1
0
Π
2Π
3
3
Π
4Π
5Π
3
3
0
2Π
x
From the 3D-plot and the contour plot, it seems
that there is a local maximum at
5π
,
point at (π, π), and a local minimum at 5π
3
3 .
π π
3, 3
fx = cos x + cos (x + y)
fy = cos y + cos (x + y)
fxx = − sin x − sin (x + y)
fyy = − sin y − sin (x + y)
fxy = − sin (x + y)
If we set fx = 0 and fy = 0, and subtract, we get cos x = cos y. Thus x = y or x = 2π − y.
If we substitute x = y into fx = 0, we get
cos x + cos 2x = 0
and since
cos 2x = 2 cos2 x − 1
we substitute to get
2 cos2 x + cos x − 1 = 0
(2 cos x − 1) (cos x + 1) = 0
so
cos x =
1
or cos x = −1
2
and thus
x=
π
5π
, x=
or x = π
3
3
, a saddle
Math 280, Sec. 14.7
2
From these x-values, we get the critical points
(π, π) ,
π π
5π 5π
, and
,
,
3 3
3 3
Now we apply the Second Partials Test to these critical points.
D (x, y) = sin x sin y + sin x sin (x + y) + sin y sin (x + y)
D (π, π) = 0
and thus the SPT does not apply. However, along the line y = x we have
f (x, x) = 2 sin x + sin 2x
= 2 sin x + 2 sin x cos x
= 2 sin x (1 + cos x)
Now f (x, x) > 0 for 0 < x < π. Also, f (x, x) < 0 for π < x < 2π. So in any neighborhood of (π, π)
there are points where the function is positive and points where the function is negative. Thus, there
is a saddle point at (π, π).
Furthermore,
D
π π 9
,
= >0
3 3
4
and
fxx
So f
π π
3, 3
=
√
3 3
2
π π
√
,
=− 3<0
3 3
is a local maximum.
Finally,
D
5π 5π
,
3 3
5π 5π
,
3 3
=
9
>0
4
and
fxx
So f
5π 5π
3 , 3
√
=
√
3>0
= − 3 2 3 is a local minimum.