Math 1920
Group Work Problems
11 Oct 2012
Very Basic Problems: If you get stuck on these, you need to review immediately.
1. In the following, there will be a function f (x, y) and an equation g(x, y) = 0, and you
are asked to find the maximum and minimum values of f (x, y) subject to the constraint
that x and y must satisfy g(x, y) = 0. In each case, first DRAW A PICTURE of the
curve g(x, y) = 0, and draw some level curves of the function f (x, y).
a) f (x, y) = 2x + 3y, g(x, y) = x2 + y 2 − 4
b) f (x, y) = x2 + y 2 , g(x, y) = 2x + 3y − 6.
Answer: For finding a maximum, say, the idea is to find a point (x0 , y0 ) at
which the only way to increase f is to move off of the curve g(x, y) = 0. This
implies (see the textbook) the condition that at (x0 , y0 ), the gradient of f is
a scalar multiple of the gradient of g. A similar statement holds for minima.
a) (∇f )(x, y) = (2, 3) and (∇g)(x, y) = (2x, 2y), so ∇f = λ∇g implies 2 = 2λx
and 3 = 2λy, or λ = 1/x = 3/(2y). This equation is satisfied whenever
x = 2y/3. To find the points on x = 2y/3 that satisfy the constraint
2
equation, we solve g(2y/3, y) = 0: (2y/3)2 + y 2 = 4 implies y√
(4/9 + 1) = 4
2
13/3. With
which
implies
y
=
4
·
13/9
which
finally
implies
y
=
±2
√
√
√
√
√
√
f ( 13, 2 13/3) = 4 13 and f (− 13, −2 13/3) = −4 13,√we see the maximum and minimum of f under the constraint are ±4 13.
b) (∇f )(x, y) = (2x, 2y), (∇g)(x, y) = (2, 3), and we know from a) that the two
are parallel only at points on the line x = 2y/3. In order to satisfy the
constraint, we need 4y/3 + 3y = 6, i.e. y(4/3 + 9/3) = 6, i.e. y(13/3) = 6,
i.e. y = 18/13. The reason we get one point is the equation is asking us
to find the point on g(x, y) = 0 with smallest squared magnitude (which
will also be the point on g(x, y) = 0 with smallest magnitude). Thus,
f (12/13, 18/13) is the minimum of f subject to g.
R1R3 1
2. Compute 0 2 (x+4y)
3 dx dy.
Answer: First,
Z
2
3
3
1
(x + 4y)−2 2
−2
1 =
(3 + 4y)−2 − (2 + 4y)−2 .
−2
(x + 4y)−3 dx =
Second,
Z 1
Z 1
Z 1
1 1
1
−2
−2
−2
(3 + 4y) − (2 + 4y)
dy =
(3 + 4y) dy −
(2 + 4y)−2 dy.
−2
−2
−2
0
0
0
1
1 1
−1 1
−2
(3
+
4y)
(3
+
4y)
dy
=
= − 41
4
−1
0
0
1
(2 + 4y)−1 = − 1 1 − 1 = 1 1 , so
Now,
1 1
4 −1
R1
4
0
Z
6
1Z
0
2
3
2
1
7
−
1
3
, and
R1
0
(2 + 4y)−2 dy =
43
1
1 11
1 1 1 1
−
−
3 dx dy =
−2 −4 7 3
−2 4 3
(x + 4y)
1 1 1 1
=
− +
8 7 3 3
1
= .
56
Similar to homework problems:
3. Find the point (a, b) on the graph of y = ex where the value ab is as small as possible.
(First, think about what you are trying to optimize and what the constraint is.)
Answer: We’re trying to minimize xy under the constraint y = ex . The
gradient of xy is (y, x) and the gradient of ex − y is (ex , −1), and the gradients
are parallel when y = λex and x = −λ for some scalar λ. But this means
y = −xex , which only lies on y = ex at x = −1. Therefore, the minimum
we seek occurs at the point (−1, e−1 ) and is −e−1 . [Note: More simply, if
y is constrained to be ex , then xy = xex ; therefore, minimizing xy subject
to y = ex is the same as minimizing xex with respect to x. Doing the
latter, we set the derivative of xex equal to zero to find the critical point:
ex + xex = ex (1 + x) = 0 implies x = −1, which concurs with the more drawn
out solution above.]
R3R1
4. Evaluate I = 1 0 yexy dy dx. You will need integration by parts and the formula
Z
ex x−1 − x−2 dx = x−1 ex + C.
Recall that a consequence of Fubini’s theorem is that the order of integration can be
switched if the integrand is a continuous function, which yexy is. Evaluate I again with
the order of integration reversed.
1
R1
R1
Answer: 0 yexy dy = xy exy 0 − x1 0 exy dy = x1 ex − x12 (ex − 1) = ex (x−1 − x−2 ) + x−2 .
Therefore,
Z 3Z 1
Z 3
Z 3
xy
x
−1
−2
ye dy dx =
e x −x
dx +
x−2 dx
1
0
1
1
3
1 −1 3
= x−1 ex 1 +
x 1
−1
1 3
1
= e −e−
−1
3
3
1 3
=
e − 1 − (e − 1) .
3
2
1
3
R3
R1
On the other hand, 1 yexy dx = exy 1 = e3y − ey , and 0 (e3y − ey ) dy = 31 e3y 0 −
1
ey 0 = 31 (e3 − 1) − (e − 1).
5. The cylinder x2 + y 2 = 1 intersects the plane x + z = 1 in an ellipse. Find the point
on the ellipse that is farthest from the origin.
Answer: We have two constraint equations, g1 (x, y, z) = x2 + y 2 − 1 and
g2 (x, y, z) = x + z − 1. Because the point on the ellipse with largest magnitude is also the point on the ellipse with the largest squared magnitude,
and because I don’t want to deal with square roots, the function I choose
to maximize is f (x, y, z) = x2 + y 2 + z 2 . Now, it is stated in the book (and
explained pretty well on the Wikipedia article for Lagrange multipliers)
that at a maximum or minimum of f subject to the constraints, the gradient of f is a linear combination of the gradients of g1 and g2 . That is,
we seek points that satisfy (∇f )(x, y, z) = λ1 (∇g1 )(x, y, z) + λ2 (∇g2 )(x, y, z) for
some λ1 and λ2 . Now, (∇f )(x, y, z) = (2x, 2y, 2z), (∇g1 )(x, y, z) = (2x, 2y, 0) and
(∇g2 )(x, y, z) = (1, 0, 1). Thus, (∇f )(x, y, z) = λ1 (∇g1 )(x, y, z) + λ2 (∇g2 )(x, y, z)
implies 2x = λ1 2x + λ2 , 2y = λ1 2y, and 2z = λ2 . Now, the second equation is
satisfied only if λ1 = 1 or y = 0. If λ1 = 1, the first equation implies λ2 = 0,
which in turn implies (via the third equation) z = 0. Thus, one critical
point of f subject to g1 and g2 is (1, 0, 0). If y = 0, then g1 (x, 0, z) = 0 implies
x = ±1. Furthermore, g2 (1, 0, z) = 0 implies z = 0 and g2 (−1, 0, z) = 0 implies
z = 2. Therefore, the two critical points of f subject to g1 and g2 are (1, 0, 0)
and (−1, 0, 2). Since f (1, 0, 0) = 1 and f (−1, 0, 2) = 5, (−1, 0, 2) is the point on
the constraint curve which is farthest from the origin.
Problems that are slightly more challenging and/or illuminate abstract
ideas:
6. Find a constraint equation g(x, y) = 0 and a continuous function f (x, y) such that f
does not achieve its minimum or maximum value on the set of (x, y) satisfying the
constraint.
Answer: Obviously the constraint curve must be unbounded (see the extreme value theorem). For example, if our constraint curve is y = x and
the [continuous] function f is f (x, y) = x, then it is easy to see that f never
achieves a maximum or a minimum on y = x.
R 1 R 1 x2 −y2
R 1 R 1 2 −y2
dx
dy
and
dy dx.
7. Evaluate 0 0 (xx2 +y
2
2)
0 0 (x2 +y 2 )2
Answer: This is in the Wikipedia article on Fubini’s theorem. The answers
are ±π/4 (not equal!!).
3