Math 2443 Homework #5
Solutions
Section 15.7.
Find the local maximum and minimum values and saddle points of the function.
2
2
8. f (x, y) = e4y−x −y .
Solution: We have
fx (x, y) = −2xe4y−x
2 −y 2
,
fy (x, y) = (4 − 2y)e4y−x
2 −y 2
.
We know that (x, y) is a critical point of f if fx (x, y) = 0 and fy (x, y) = 0. Since
the exponential is never zero, this means that x = 0 and 4 − 2y = 0. Hence, (0, 2)
2
2
is the only critical point. We also have fxx (x, y) = (−2 + 4x2 )e4y−x −y , fyy (x, y) =
2
2
2
2
(−2 + (4 − 2y)2 )e4y−x −y and fxy (x, y) = (−2x)(4 − 2y)e4y−x −y . Hence,
D(0, 2) = fxx (0, 2)fyy (0, 2) − (fxy (0, 2))2 = 4e8 > 0.
Since fxx (0, 2) = −2e4 < 0, by the Second Derivative Test, the function has a local maximum at (0, 2) and f (0, 2) = e4 is a local maximum value.
12. f (x, y) = xy + x1 + y1 .
Solution: We have fx (x, y) = y − x12 and fy (x, y) = x − y12 . The critical points are thus
given by x2 y = 1 and y 2 x = 1, and thus (1, 1) is the only critical point. We calculate
fxx (x, y) = x23 , fyy (x, y) = y23 and fxy (x, y) = 1. Thus,
D(x, y) = fxx (x, y)fyy (x, y) − (fxy (x, y))2 =
4
−1
x3 y 3
so D(1, 1) = 3 > 0 and fxx (1, 1) = 2 > 0. By the Second Derivative Test, the function
therefore has a local minimum at (1, 1) and f (1, 1) = 3 is a local minimum value.
16. f (x, y) = ey (y 2 − x2 ).
Solution: We have fx (x, y) = −2xey and fy (x, y) = ey (y 2 − x2 + 2y). Thus, (x, y) is a
critical point if x = 0 and y 2 + 2y = 0. The critical points are therefore (0, 0) and (0, −2).
Now, we calculate fxx (x, y) = −2ey , fyy (x, y) = (y 2 −x2 +4y+2)ey and fxy (x, y) = −2xey .
We have fxx (0, 0) = −2 < 0 and fxx (0, −2) = −2e−2 < 0 and
D(0, 0) = (−2)(2) = −4 < 0,
1
D(0, −2) = 4e−2 > 0.
2
Solutions
This means that, by the Second Derivative Test, the function has a local maximum at
(0, −2) and the local maximum value f (0, −2) = 4e−2 . The point (0, 0) is a saddle point.
Find the absolute maximum and minimum values of f on the set D.
30. f (x, y) = 3+xy−x−2y, D is the closed triangular region with vertices (1, 0), (5, 0) and (1, 4).
Solution: (a) We first find the critical points (points at which fx = 0, fy = 0). For the
function f we have fx (x, y) = y − 1 and fy (x, y) = x − 2. Hence, the only critical point is
(2, 1) and f (2, 1) = 1.
(b) Next, we find the extreme values of the function on the boundary of the given region.
The triangular region consists of three line segments. Let L1 be the segment joining (1, 0)
and (5, 0), let L2 be the segment joining (1, 4) and (5, 0) and L3 the segment joining (1, 4)
and (1, 0).
(i) On L1 we have y = 0 thus the function is f (x, 0) = 3 − x. Since x lies between 1
and 5, the minimum value of f on L1 is f (5, 0) = −2 and the maximum value is
f (1, 0) = 2.
(ii) The equation of the line which contains L2 is y = −x + 5. Thus, the function f on
L2 is f (x, −x + 5) = −x2 + 6x − 7. To find the max/min values of this function, we
complete the square to get f (x, −x + 5) = −(x − 3)2 + 2. Then, the function has
a minimum when x = 1 and x = 5, that is f (1, 4) = −2 = f (5, 0). The maximum
value is when x = 3. Thus f (3, 2) = 2 is the maximum value.
(iii) On L3 we have x = 1, so f (1, y) = 2 − y. The maximum and minimum values of
this function on L3 are f (1, 0) = 2 and f (1, 4) = −2 respectively.
Hence, comparing all the values, we see that the absolute maximum and absolute minimum values of the function are 2 and −2 respectively.
34. f (x, y) = xy 2 , D = {(x, y)| x ≥ 0, y ≥ 0, x2 + y 2 ≤ 3}.
Solution: The procedure is the same as in the previous problem.
(a) We have fx = y 2 and fy = 2xy. Hence fx = 0 and fy = 0 if y = 0. Technically, x can
have any value but remember
that we must restrict ourselves to the region D in which, if
√
y = 0, then 0 ≤ x < 3. For all these critical points, y = 0 so the value of the function is
zero at each of the critical points.
(b) Now, we calculate the extreme values of the function on the boundary x2 + y 2 = 3.
We substitute y 2 = 3 − x2 into the function to get
p
f (x, 3 − x2 ) = x(3 − x2 ).
This√function has a maximum on the boundary when x = 1, and the maximum value is
f (1, 2) = 2. Likewise, the minimum value of f on the boundary is −2.
Math 2443 Homework #5
3
Comparing values from (a) and (b), the absolute maximum value of f on D is 2 and the
absolute minimum value is −2.
40. Find the point on the plane x − y + z = 4 that is the closest to the point (1, 2, 3).
Solution: Take p
an arbitrary point (x, y, z) on the given plane. The distance of this point
from (1, 2, 3) is (x − 1)2 + (y − 2)2 + (z − 3)2 . Since it will be easier to minimize the
square of the distance, we let
f (x, y, z) = (x − 1)2 + (y − 2)2 + (z − 3)2
Since (x, y, z) lies on the given plane, we also have z = 4 − x + y. Substituting this in the
function, we can define a new function
g(x, y) = f (x, y, 4 − x + y) = 2x2 − 4x − 2y − 2xy + 2y 2 + 6.
So the problem reduces to finding the minimum value of g. We compute
gx = 4x − 4 − 2y,
gy = 4y − 2 − 2x.
Setting each of the equations to zero and solving, we see that ( 53 , 43 ) is the only critical
point. Notice that gxx = 4 > 0, gyy = 4 and gxy = −2. Hence D = 4(4) − (−2)2 = 12 > 0.
Hence, the function has a local minimum at ( 53 , 34 ). For these values of x and y, we have
z = 4 − 53 + 34 = 11
3 .
In conclusion, the point on the plane x − y + z = 4 that is closest to (1, 2, 3) is 35 , 43 , 11
3 .
44. Find three positive numbers whose sum is 12 and the sum of whose squares is as
small as possible.
Solution: Let the positive numbers be x, y and z. Then, we are given that x + y + z = 12
and we are asked to minimize the function
f (x, y, z) = x2 + y 2 + z 2 .
Define a function g such that
g(x, y) = f (x, y, 12 − x − y) = x2 + y 2 + (12 − x − y)2 = 2x2 + 2y 2 − 24x + 24y − 2xy + 144.
We now have to find the minimum value of g. This is very similar to the previous problem
so I will let you complete it.
46. Find the dimensions of the box with volume 1000 cm2 that has minimal surface area.
4
Solutions
Solution: Let the dimensions of the box be x, y and z. We are given that xyz = 1000.
The surface area is given by f (x, y, z) = 2(xy + yz + xz) = 2xy + 2(x + y)z. Define a
function g by
1000
1000
2000 2000
g(x, y) = f x, y,
= 2xy + 2(x + y)
= 2xy +
+
.
xy
xy
y
x
The problem asks to minimize this function. We have
gx = 2y −
2000
,
x2
gy = 2x −
2000
y2
and the critical points are given by yx2 = 1000 and xy 2 = 1000. Since x, y 6= 0 (the
dimensions can’t be zero), dividing the first equality by the second, we get x = y and
thus x = y = 10. From this, we also have z = 1000
xy = 10. Let’s use the Second Derivative
4000
4000
Test now. We calculate gxx = x3 , gyy = y3 and gxy = 2. Hence, fxx (10, 10) > 0 and
D(10, 10) > 0. Thus, g has a minimum at (10, 10) and consequently, the dimensions of
the box with volume 1000 and minimal surface area must be x = y = z = 10.