Recall from Theorem 4.4.4 that if a continuous function of one variable has exactly one relative extremum on an interval, then that relative extremum is an absolute extremum on the interval. This exercise shows that this result does not extend to functions of two variables. (a) Show that f x , y = 3 x e y − x 3 − e 3 y has only one critical point and that a relative maximum occurs there. (See the accompanying figure.) (b) Show that f does not have an absolute maximum.
Recall from Theorem 4.4.4 that if a continuous function of one variable has exactly one relative extremum on an interval, then that relative extremum is an absolute extremum on the interval. This exercise shows that this result does not extend to functions of two variables. (a) Show that f x , y = 3 x e y − x 3 − e 3 y has only one critical point and that a relative maximum occurs there. (See the accompanying figure.) (b) Show that f does not have an absolute maximum.
Solution Summary: The author demonstrates that the function f(x,y) has only one critical point and has a relative maximum.
Recall from Theorem 4.4.4 that if a continuous function of one variable has exactly one relative extremum on an interval, then that relative extremum is an absolute extremum on the interval. This exercise shows that this result does not extend to functions of two variables.
(a) Show that
f
x
,
y
=
3
x
e
y
−
x
3
−
e
3
y
has only one critical point and that a relative maximum occurs there. (See the accompanying figure.)
(b) Show that f does not have an absolute maximum.
Formula Formula A function f(x) attains a local maximum at x=a , if there exists a neighborhood (a−δ,a+δ) of a such that, f(x)<f(a), ∀ x∈(a−δ,a+δ),x≠a f(x)−f(a)<0, ∀ x∈(a−δ,a+δ),x≠a In such case, f(a) attains a local maximum value f(x) at x=a .
Write the given third order linear equation as an equivalent system of first order equations with initial values.
Use
Y1 = Y, Y2 = y', and y3 = y".
-
-
√ (3t¹ + 3 − t³)y" — y" + (3t² + 3)y' + (3t — 3t¹) y = 1 − 3t²
\y(3) = 1, y′(3) = −2, y″(3) = −3
(8) - (888) -
with initial values
Y
=
If you don't get this in 3 tries, you can get a hint.
Question 2
1 pts
Let A be the value of the triple integral
SSS.
(x³ y² z) dV where D is the region
D
bounded by the planes 3z + 5y = 15, 4z — 5y = 20, x = 0, x = 1, and z = 0.
Then the value of sin(3A) is
-0.003
0.496
-0.408
-0.420
0.384
-0.162
0.367
0.364
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