|
5) | The maximum, 14, occurs where x = 7 and y = 0. The minumum, -14, occurs where x = -7 and
y = 0. |
|
6) | The problem is to minimize f(x,y) = 60 x + 100 y subject to the constraint
500 x0.4 y 0.8 = 104.
The minimum, 2400 (5/6)4/5 occurs at x = 20(5/6)(5/6)2/3 and y = 24(5/6)(5/6)-1/3. |
|
7) | Denote the radius of the cylinder by r and the height by h. You are being asked to minimize
 (r2 + 2 h r) subject to the constraint r2 h = 50.
The minimum occurs where r = (50/)1/3 and
h = (50/)1/3. |
|
8) | The point has coordinates (101)-1/2 (8, 27, 4). One way to find this point is to realize that the
gradient of x2/4 + y2/9 + z2/4 is normal to the plane 2x + 3y + z at the points with minimum
and maximum distance. |
|
9) | Suppose that corners of the box have coordinates (±x, ±y, ±z), where x, y and z are positive.
These corners will lie on the ellipse (otherwise, the box could be made bigger). Thus, you are
asked to maximize the volume, 8xyz, subject to the constraint x2/4 + y2/9 + z2/4 = 1. The
maximum occurs where x = 2 3-1/2, y = 31/2,
z = 2 3-1/2. |
|
10) | An open-top rectangular box of side length x, y, and z (height) has volume xyz and surface area
that is equal to xy + 2(xz + yz) = 36. The maximum volume is z=3 1/2
for x = y = 2 31/2
and z =31/2. Meanwhile, an open-top cylindrical box with radius r and height z has volume
r2h and surface area r2 + 2rh = 36.
The maximum volume here is 24 (3/)1/2 which occurs, when
r = h = (12/)1/2.
Thus, the cylindrical box has 8 () -1/2 times as much volume as the rectangular
one. |
|
11) | You are being asked to minimize the function 80x + 25y + 15z subject to the constraint that
300x2/5y1/2z1/10 = 12,000.
The minimum occurs for x = 10 (768)1/10, y = 40 (768)1/10 and z =
40 (786)1/10/3. |
|
12a) | You are being asked to minimuze the function 35x + 16y subject to
500x7/10 y1/2 = 40,000. The minimum occurs at x = 32, y = 50.
b) You are being asked to maximize 500x7/10y1/2 subject to 35x + 16y = 4800. The maximum
occurs at x = 80, y = 125. |
|
13a) | At ten months, x = 100, y = 125 so the money being spent is 100x + 120y = 25,000 and the
production is 300x1/2 y1/3 = 15,000. |
|
13b) | You are asked to evaluate  P at x = 100 and y = 125.
The answer is 75. |
13c) | You are asked to evaluate  P at x=100 and y=125.
The answer is 40. |
|
13d) | You are asked to maximize 300x1/2y1/3 subject to
100x + 120y = 25,000. The maximum
occurs at x = 150, y = 250/3 and equals 7500 25/6 31/6. |
|
13e) | You are asked to evaluate P at x = 150, y = 250/3.
The answer is 25 25/6 31/6. |
