MATH 114, 2009 Quiz # 8 Solution Thursday
Name: Section: 207, 208
1. Consider the following function f(x, y, z) = x2+ 2y2−z.
(1) For any point (x, y, z), what is the maximum directional derivative?
Ans. Note that for any unit vector u, the directional derivative in udirection is
Duf=Of•u
and it obtains the maximum value when uhas Ofdirection, i.e., when
u=Of
|Of|,
and in that case,
Duf=Of•Of
|Of|=Of•Of
|Of|=|Of|2
|Of|=|Of|.
For given f(x, y, z) = x2+ 2y2−z,Of=<2x, 4y , −1>, hence the maximum directional
derivative is
|Of|=p4x2+ 16y2+ 1.
(2) Now consider the plane x+ 2y+z= 1. Using the Lagrange multipliers method, find
all the extreme values of f(x, y, z) on this plane.
At what point does f(x, y, z) have its minimum value on the plane?
Ans. Let g(x, y, z) = x+ 2y+z. Then g(x, y, z) = 1 is the given constraint. We find
all x, y, z and λvalues satisfying
Of=λOg,
x+ 2y+z= 1.
Since Of=<2x, 4y, −1>,Og=<1,2,1>, from the first equation, we get
<2x, 4y, −1>=λ < 1,2,1>⇒
2x=λ
4y=λ
−1 = λ⇒
x=−1/2
y=−1/2
z= 5/2
The last zvalue comes from the constraint equation by plugging in x=−1/2, y =−1/2.
From the Lagrange multipliers method, we only get one extreme value at (−1/2,−1/2,5/2)
which is
f(−1/2,−1/2,5/2) = −1
22
+ 2 −1
22
−5
2=−7
4.
To check if it is the minimum value, we can just find another function value at a point on
the constraint which is bigger than −7/4. Let’s say we choose (0,0,1). (It surely satisfies
g(x, y, z) = 1 equation). At that point, the function value is
f(0,0,1) = −1
