Name
MATH 253
Sections 501-503
ID
Section
EXAM 1
Fall 1998
P. Yasskin
Multiple Choice: (3 points each)
1-8
/24
9
/6
10
/10
11
/10
Problems 1 – 3: Find the plane tangent to the graph of the equation z lnŸy " z lnŸx e 2
at the point Ÿx, y, z Ÿ1, e, e 2 . Write the equation of the plane in the form z Ax By C
and find the values of A, B and C in problems 1, 2 and 3:
Solution: FŸx, y, z z lnŸy " z lnŸx METHOD 1:
FŸ1, e, e 2 Ÿ"e 2 , e, 1 F " z , z , lnŸy " lnŸx N
X Ÿx, y, z P Ÿ1, e, e 2 x y
X N
P
Plane:
N
"e 2 x ey z "e 2 1 ee e 2 e 2
METHOD 2:
F x Ÿx, y, z " xz
F y Ÿx, y, z yz
F z Ÿx, y, z lnŸy " lnŸx z e 2 x " ey e 2
F x Ÿ1, e, e 2 "e 2
F z Ÿ1, e, e 2 1
F y Ÿ1, e, e 2 e
F x Ÿ1, e, e 2 Ÿx " 1 F y Ÿ1, e, e 2 Ÿy " e F z Ÿ1, e, e 2 Ÿz " e 2 0
" e 2 x e 2 ey " e 2 z " e 2 0
"e 2 Ÿx " 1 eŸy " e 1Ÿz " e 2 0
B "e
C e2
In either case:
A e2
z e 2 x " ey e 2
1. A a. 1
b. e
c. e 2 T correctchoice
d. "e
e. "e 2
2. B a. 1
b. e
c. e 2
d. "eT correctchoice
e. "e 2
3. C a. 1
b. e
c. e 2 T correctchoice
d. "e
e. "e 2
1
x"y
at
x2 y2
the point Ÿx, y Ÿ2, 1 . Write the equation of the plane in the form z Ax By C and
find the values of A, B and C in problems 4, 5 and 6:
x"y
fŸ2, 1 2 " 1 1
Solution: fŸx, y 2
5
41
x y2
2
2
Ÿx y 1 " Ÿx " y 2x
Ÿ4 1 1 " Ÿ2 " 1 4
1
f x Ÿx, y f x Ÿ2, 1 2
2
2 2
25
Ÿ4 1 Ÿx y 2
2
Ÿx y Ÿ"1 " Ÿx " y 2y
Ÿ4 1 Ÿ"1 " Ÿ2 " 1 2
"7
f y Ÿx, y f y Ÿ2, 1 2
2
2 2
25
Ÿ4 1 Ÿx y z fŸ2, 1 f x Ÿ2, 1 Ÿx " 2 f y Ÿ2, 1 Ÿy " 1 1 1 Ÿx " 2 " 7 Ÿy " 1 5
25
25
7
1
1
7
1
7
2
"
2
x"
y x"
y
5
5
25
25
25
25
25
25
7
2
1
B"
C
So
A
5
25
25
Problems 4 – 6: Find the plane tangent to the graph of the function fŸx, y 4. A a.
b.
c.
d.
e.
5. B a.
b.
c.
d.
e.
6. C 1 T correctchoice
25
7
25
"1
5
" 1
25
" 7
25
1
25
7
25
"1
5
" 1
25
" 7 T correctchoice
25
a. 2 T correctchoice
5
b. 1
5
c. 0
d. " 1
5
2
"
e.
5
2
7. The dimensions of a closed rectangular box are measured as 80 cm, 60 cm and 50 cm
with a possible error of 0.2 cm in each dimension. Use differentials to estimate the
maximum error in the calculated surface area of the box.
Solution: A 2xy 2xz 2yz
dA A dx A dy A dz Ÿ2y 2z dx Ÿ2x 2z dy Ÿ2x 2y dz
y
z
x
Ÿ120 100 .2 Ÿ160 100 .2 Ÿ160 120 .2 152
38 cm 2
95 cm 2
144 cm 2
148 cm 2
e. 152 cm 2 T correctchoice
a.
b.
c.
d.
8. Consider the function fŸx, y 3Ÿx y 3 . Find the set of all points Ÿx, y where
f 0.
f 9Ÿx y 2 , 9Ÿx y 2 0
when
which is the line
Solution: xy 0
which may be parametrized as
y "x
Ÿx, y Ÿt, "t .
a. The point Ÿx, y Ÿ0, 0 .
b. The point Ÿx, y Ÿ1, "1 .
c. The line Ÿx, y Ÿt, "t .T correctchoice
d. The line Ÿx, y Ÿt, t .
e. The circle x 2 y 2 1 .
9
9. (6 points) Suppose p pŸx, y , while x xŸu, v and y yŸu, v . Further, you know the
following information:
xŸ1, 2 3
p
Ÿ1, 2 7
x
x Ÿ1, 2 11
u
y
Ÿ1, 2 15
u
yŸ1, 2 4
pŸ1, 2 5
pŸ3, 4 6
p
p
p
Ÿ1, 2 8
Ÿ3, 4 9
Ÿ3, 4 10
y
x
y
x Ÿ1, 2 12
x Ÿ3, 4 13
x Ÿ3, 4 14
v
u
v
y
y
y
Ÿ1, 2 16
Ÿ3, 4 17
Ÿ3, 4 18
v
u
v
p
. Then use it and the above information to compute
Write out the chain rule for
v
p
Ÿ1, 2 .
v
p
y
x p
Solution:
p x p y p
x v
y v
x ŸxŸu,v ,yŸu,v v
y x u,v ,y u,v v
v
Ÿ Ÿ
Ÿ
p
y
p
x Ÿ1, 2 p
Ÿ1, 2 Ÿ1, 2 v
x ŸxŸ1,2 ,yŸŸ1,2 v
y x 1,2 v ,y 1,2 v
Ÿ ŸŸ
ŸŸ
p Ÿ3, 4 x Ÿ1, 2 p Ÿ3, 4 y Ÿ1, 2 9 12 10 16 268
v
x
y
v
3
10. (10 points) Find all critical points of the function fŸx, y xy 2 3x 2 " 3y 2 " 2x 3 and
classify each as a local maximum, a local minimum or a saddle point.
f y 2xy " 6y 2yŸx " 3 0
Solution: f x y 2 6x " 6x 2 0
From f y :
y0
or
x3
So: x 0
or
x 1.
6x " 6x 2 6xŸ1 " x 0
Case: y 0:
From f x :
2
2
y 36
So: y 6
or
y "6.
Case: x 3:
From f x :
y 18 " 54 0
So the critical points are:
Ÿ0, 0 Ÿ1, 0 Ÿ3, 6 Ÿ3, "6 f xx 6 " 12x
To classify:
f yy 2x " 6
f xy 2y
x
y
f xx
f yy
f xy
D
Classification
0
0
6
"6
0
"36 0
saddle
1
0
"6 0 "4
0
24 0
local maximum
3
6
"30
0
12
"144 0
saddle
3 "6
"30
0
"12 "144 0
saddle
D f xx f yy " f xy2
11. (10 points) Find the dimensions and volume of the largest rectangular box in the first
octant with three faces in the coordinate planes and one vertex in the plane
3x 2y z 6.
Solution: Maximize
V xyz
subject to
g 3x 2y z 6.
METHOD 1:
Eliminate a variable:
z 6 " 3x " 2y
V xyŸ6 " 3x " 2y 6xy " 3x 2 y " 2xy 2
V y 6x " 3x 2 " 4xy xŸ6 " 3x " 4y 0
V x 6y " 6xy " 2y 2 yŸ6 " 6x " 2y 0
To have a non-zero volume, we must have x p 0 and y p 0.
So
6 " 6x " 2y 0
6 " 3x " 4y 0
or
6x 2y 6
or
12x 4y 12
3x 4y 6
3x 4y 6
Subtracting, we have: 9x 6 or x 2 . So 3 2 4y 6 or 4y 4 or y 1.
3
3
2
So z 6 " 3
" 2Ÿ1 2
3
So the dimensions are x 2 , y 1 and z 2. The volume is V 4 .
3
3
g Ÿ3, 2, 1 V Ÿyz, xz, xy METHOD 2:
Lagrange Multipliers:
g
V 5
Ÿyz, xz, xy 5Ÿ3, 2, 1 Solve yz 35 xz 25 xy 5 and 3x 2y z 6 for x, y, z and 5.
5 xy yz xz
3z 6
x z
y z
3 z 2 z z 6
3
3
2
2
3
2
z2
x 2
y1
The volume is V 4 .
3
3
4