MATHEMATICS 3202
ASSIGNMENT 7 - SOLUTION
Due on Monday, March 7, 2016; Submitted to Department Exercise Box 33
Name
M.U.N. Number
p
1. Show that f (x, y) = x2 + y 2 has one critical point P and that f is non-differentiable at P . Does f take on a
minimum, maximum, or saddle point at P ?
Solution [p.843 ex. 6; SM p.755] Since f ≥ 0 and f (0, 0) = 0 is an absolute minimum value. To find the critical
point of f we first find
y
x
& fy (x, y) = p 2
fx (x, y) = p 2
2
x +y
x + y2
which do not exist at (0, 0) and fx (x, y) = 0 = fy (x, y) has no solutions, the only critical point is (0, 0) - a point
where f is non-differentiable (and is the absolute minimum).
2. Show that f (x, y) = x2 has infinitely many critical points (as a function of two variables) ant that the Second
Derivative Test fails for all of them. What is the minimum value of f ? Does f (x, y) have any local maxima?
Solution [p.844 ex. 24; SM p.765] First if we solve for critical points we get fx (x, y) = 2x, and fy (x, y) = 0.
This setting each equal to zero only yields x = 0 and y can be any real number. The list of critical points is
{(0, r) : r ∈ (−∞, ∞)}. Now computing the second-order partials for the discriminant we get fxx (x, y) = 2,
fxy (x, y) = 0, fyy (x, y) = 0. Thus, D = 0. This means that the Second Derivative Test is inconclusive for every
critical point, it fails. Finally, this function does have a minimum value of 0 since the smallest any square can be
is 0. Since x can get arbitrarily large, this function has no maximum value, and no local maxima.
3. Find a continuous function that does not have a global maximum on the domain D = {(x, y) : 0 < x < 1, 0 < y <
1}. Explain why this does not contradict Theorem on Existence and Location of Global Extrema.
Solution [p.844 ex. 34; SM p.769] Consider f (x, y) = 2−1 x2 and take fx = x, fy = 0, and fxx = 1, fyy = 0, fxy = 0.
This implies D = 0. So, the Second Derivative Test is going to be inconclusive.
Considering this function over the domain, D = {(x, y) : 1 ≥ x + y ≥ 0}, we see that f (x, y) = 2−1 x2 is in the strip
formed between to the two lines y = −x and y = 1 − x. We can make f (x, y) = 2−1 x2 arbitrarily large within this
region. In fact, we can see that limx→−∞ f (x, y) is arbitrarily large. This does not contradict the theorem in the
text, because D is an bounded domain, in that for any integer n, we can see that the open interval (−n, n + 1/2)
is contained in this region.
4. Find the extreme values of f (x, y) = x2 + 2y 2 subject to the constraint g(x, y) = 4x − 6y = 25.
Solution [p.852 ex. 2; SM p.791]
Since ∇f = h2x, 4yi and ∇g = h4, −6i. The Lagrange equations are thus ∇f = λ∇g, i.e., 2x = 4λ and 4y = −6λ.
This gives 4y = −3x. Since 4x − 6y = 25, we have x = 50/17 and hence y = −(75)/(34). We conclude that
(50/17, −75/34) is a unique critical point. The extreme value of f is: (50/17)2 + 2(75/34)2 .
5. In the following find the minimum and maximum values of the function subject to the given constraint: i) f (x, y) =
2x+3y, x2 +y 2 = 4; ii) f (x, y) = 4x2 +9y 2 , xy = 4; iii) f (x, y) = x2 y+x+y, xy = 4; iv) f (x, y) = x2 y 4 , x2 +2y 2 = 6;
v) f (x, y, z) = x2 − y − z, x2 − y 2 + z = 0.
Solution [p.853 ex.4,6,8,10,12; SM p.793-799] i) Since ∇f = h2, 3i and ∇g = h2x, 2yi, the Lagrange Condition is
−1
2
2
h2, 3i = λh2x, 2yi. So, x 6= 0 6= y. This
√ gives λ = x =√3/(2y). Since g(x, y) = x + y = 4, we solve for x and
y via this condition, and get x = ±4/ 13 and y = ±6/ 13. Extreme points may occur also where ∇g = h0, 0i.
However, (0, 0) is not on the constraint.
√
√
√
√
Therefore the maximum is f (4/ 13, 6/ 13) ≈ 7.21, and the minimum is f (−4/ 13, −6/ 13) ≈ −7.21.
√ p
√
√ p
p
ii) Similarly,
consequently, f ( 6, 2 2/3) = 48 =
√
p the critical points are ( 6, 2 2/3) and (− 6, −2 2 2/3) and
f (− 6, −2 2/3). On the constraint y = 4/x, thus f (x, y) = 4x + 144x−2 → ∞ as |x| → ∞. This means that
f (x, y) has a global minimum of 48 on (−∞, ∞).
iii) Under xy = 4, we have f (x, y) = 4x + x + 4/x → ∞(or − ∞) as x → 0 + (or 0−). So, there are no minimum
and maximum values of f (x, y) under the constraint.
iv) We find the extreme values of f (x, y) = x2 y 4 on the constraint g(x, y) = x2 + 2y 2 − 6 = 0. Using the Lagrange
Equation to get the following critical points
√
√
√ √
√
√
√ √
( 2, − 2), ( 2, 2), (− 2, − 2), (− 2, 2).
Extreme values can occur also at the point where ∇g = h0, 0i - however this point does not lie on the constraint.
So the extreme value is
√
√
√ √
√
√
√ √
f ( 2, − 2) = f ( 2, 2) = f (− 2, − 2) = f (− 2, 2) = 8.
Recall that there are critical points with x = 0 or y = 0 at which the value of f is zero. Since f has global extrema
on the ellipse x2 + 2y 2 = 6, we conclude that the minimum value of f on the constraint is 0 and the maximum
value is 8.
v) We show that f (x, y, z) = x2 − y − z does not have minimum and maximum values subject to the constraint
g(x, y, z) = x2 − y 2 + z = 0. Note that the curve (x, x, 0) lies on the constraint, since it satisfies the equation of
the constraint. On this curve we have f (x, y, z) = x2 − x → ∞ as |x| → ∞, and consequently, f does not have
√
a maximum value subject to the constraint. Observe that the curve (0, z, z) also lies on the constraint, and we
√
have f (x, y, z) = −(z + z) → −∞ as z → ∞, thereby getting that f does not attain a minimum value on the
constraint either.
6. Find the rectangular box of maximum volume if the sum of the lengths of the edges is 300cm.
Solution [p.853 ex. 16; SM p.801] Denote by x, y, z the dimensions of the rectangular box. Then volume of the
box is xyz. The constraint condition is g(x, y, z) = x + y + z = 300 and x, y, z ≥ 0. (one could also argue that the
sums of the lengths of the edges is 4(x + y + z) = 300 but that would give a different answer, of course, instead,
we will choose to interpret the problem with the constraint x + y + z = 300).
Write out the Lagrange Equations ∇f = λ∇g, i.e., yz = xz = xy = λ. If x = 0 or y = 0 or z = 0, then the volume
has the minimum value 0. Thus we may assume that all x, y, z are nonzero. Consequently, x = y = z. This is put
in x + y + z = 300 and gives x = y = z = 100. Therefore, the extreme value is f (100, 100, 100) = 1003 = 106 cm3 .
This must be maximum value. And 0 is the minimum value of f .
7. Evaluate
R
[1,4]×[1,3] (x
− 2y)dA.
Solution [p.870 ex. 6; SM p.869]
Z
(x − 2y)dA =
[1,4]×[1,3]
Z 4Z 3
0
(x − 2y)dydx =
1
Z 4
1
3
Z 4
1
1
(xy − y 2 ) dx = 2
(x − 4)dx = −9.
8. Evaluate the following integrals:
Z
x2 y dA; ii)
i)
Z
[−1,1]×[0,2]
y(x + 1)−1 dA; iii)
[0,2]×[0,4]
Solution [p.871 ex. 38, 40, 42; SM p.878-879]
i)
R
ii)
iii)
2
[−1,1]×[0,2] x y
R
[0,2]×[0,4] y(x
dA =
0
−1 x
+ 1)−1 dA =
3x+4y
[0,1]×[1,2] e
R
R2R1
dA =
2 ydxdy
R4R2
0
0
= 4/3.
y(x + 1)−1 dxdy = 8 ln 3.
R 1 R 2 3x+4y
3
8 −e)
dydx = (e −1)(e
.
0 1 e
12
Z
[0,1]×[1,2]
e3x+4y dA.
9. Evaluate I = 13 01 yexy dydx. You will need Integration by Parts and the formula ex (x−1 − x−2 ) dx = x−1 ex + C.
Then evaluate I again using Fubini’s Theorem to change the order of integration (that is, integrate first with
respect to x). Which method is easier?
R R
R
Solution [p.871 ex 44; SM p.880] We evaluate the inner integral with respect to y. Using integration by parts with
dv = exy dy v = y we obtain
Z
−1 xy
xy
ye dy = yx
hence
e
Z 1
−
Z
x−1 exy dy = x−1 exy (y − x−1 ) + C
yexy dy = ex (x−1 − x−2 ) + x−2 .
0
We now integrate the result with respect to x, using the given integration formula:
Z 3
1
3
3
1
1
(ex (x−1 − x−2 ) + x−2 )dx = x−1 ex − x−1 = 3−1 e3 − e + 2/3 ≈ 4.644.
The double integral is thus
Z 3Z 1
yexy dydx =
(
yexy dy)dx ≈ 4.644.
0
1
0
1
Z 3 Z 1
Using Fubini’s Theorem, we now evaluate the double integral first with respct to x and then with respect to y,
and obtain
Z 3Z 1
1
yexy dydx =
Z 1Z 3
Z 1
(e3y − ey )dy = 3−1 e3 − e + 2/3 ≈ 4.644.
0
1
0
0
yexy dx dy =
Obviously, inegrating first with respect to x and then with respect to y makes the computation much easier than
using the reversed order.
10. Evaluate
R1R1
0
0
y(1 + xy)−1 dydx. Hint: Change the order of integration.
Solution [p.871 ex 45; SM p.880]
Using Fubini’s Theorem, we get
Z 1Z 1
y(1 + xy)
0
0
−1
dydx =
Z 1Z 1
0
0
−1
y(1 + xy)
Z 1
dx dy =
0
Z 1
1
ln(1 + y)dy = 2(ln 2 − 1).
ln(1 + xy) dy =
0
0