Solution

Math 21a: Multivariable calculus
Fall 2014
Homework 18: Extrema
This homework is due Wednesday, 10/22 rsp Thursday 10/23.
1 Find the local maximum and minimum values of the function.
f (x, y) = 5xy +
5 5
+
x y
Solution:
The gradient of f is ∇f (x, y) = h5y − x52 , 5x − y52 i. Setting the
gradient equal to ~0, we solve for the critical point: (1,1). To
determine if this is a local maximum or local minimum, we use
the second derivative test. Compute:
fxx = 10
x3 ⇒ fxx (1, 1) = 10 > 0
fyy = 10
y 3 ⇒ fyy (1, 1) = 10
fxy = 5 ⇒ fxy (1, 1) = 5
D(1, 1) = fxx(1, 1)fyy (1, 1) − (fxy (1, 1))2 = 75 > 0.
Therefore, (1,1) is a local minimum and its value is f (1, 1) = 15.
2 Classify the critical points of the function
f (x, y) = 7e2y (4y 2 − x2) .
Solution:
First, we find the critical points. Compute the gradient:
∇f = h(7e2y )(−2x), (7e2y )(8y) + (4y 2 − x2)(14e2y )i.
Setting the gradient equal to ~0, we get two equations:
−14xe2y = 0
and
7e2y (8y + 8y 2 − 2x2) = 0.
We solve for the critical points and get: (0, 0) and (0, -1).
Next, we classify the critical points using the second derivative
test. Compute the matrix of second derivatives:
−14e2y
−28xe2y
2y
2y
−28xe
28e (2 + 8y + 4y 2 − x2 ))
!
At (0,0), this matrix becomes
−14 0
0
56
!
which has a negative determinant and thus (0,0) is a saddle
point.
At (0, -1), the matrix is
−14e−2
0
0
−56e−2
!
which has a positive determinant. Also, fxx(0, −1) < 0, so (0,
-1) is a local maximum.
3 Find the local maximum and minimum values and saddle point(s)
of the function.
f (x, y) = 2 sin x sin y, −π < x < π, −π < y < π
Solution:
First we find the critical points. Compute the gradient:
∇f = h2 cos x sin y, 2 sin x cos yi.
Setting the gradient equal to ~0, we get two equations:
2 cos x sin y = 0;
2 sin x cos y = 0.
From the first equation, we see that either cos x = 0 or sin x = 0.
If cos x = 0, then x = ±π/2. Going over to the second equation,
if x = ±π/2, then we must have that cos y = 0, so y = ±π/2.
If instead sin y = 0 then y = 0. Going over to the second
equation, if y = 0, then sin x = 0 so x = 0. Thus, the critical
points in the range provided are: (±π/2, ±π/2) and (0,0).
Now, we classify the points. The matrix of second derivatives is
−2 sin x sin y 2 cos x cos y
2 cos x cos y −2 sin x sin y
!
At (0, 0), we get
0 2
2 0
!
which has a negative determinant so the point (0,0) is a
saddle point.
At (π/2, π/2) and (−π/2, −π/2), we get
−2 0
0 −2
!
which has a positive determinant and fxx < 0, so the points
(π/2, π/2) and (−π/2, −π/2) are both local maximums.
Finally, at (π/2, −π/2) and (−π/2, π/2), we get
2 0
0 2
!
which has a positive determinant while fxx > 0, so the
points (π/2, −π/2) and (−π/2, π/2) are both local minimums.
4 Companies like Netflix or Hulu track movie preferences. One
can visualize preferences on parameter spaces which is the intelligence - emotion plane. Based on viewing habits, the service decides what you want to see. Your profile is a function
f (x, y). Maximizing this function allows the company to pick
movies for you. Assume that your user profile is the function
f (x, y) = −2x3 + 9x2 − 12x − y 2. Find and classify all the critical
points and especially find the local maxima of f .
Solution:
First we find the critical points. The gradient of f is ∇f =
h−6x2 + 18x − 12, −2y 2i. This equals ~0 at the points (1,0) and
(2,0).
Next, we classify the critical points. The matrix of second
derivatives is
!
−12x + 18 0
0
−2
At the point (1,0), we get
6 0
0 −2
!
which has a negative determinant, so (1,0) is a saddle point.
At the point (2,0), we get
−6 0
0 −2
!
which has a positive determinant and fxx < 0, so (2,0) is a
local maximum.
Graph theorists look at the Tutte
polynomial f (x, y) of a network. We
work with the Tutte polynomial
5
f (x, y) = x + 2x2 + x3 + y + 2xy + y 2
of the Kite network. Classify the two
critical using the second derivative test.
Remark.
The polynomial is useful:
xf (1 − x, 0)
tells in how many ways one can color the nodes of
the network with x colors and f (1, 1) tells how many
spanning trees there are.
This picture illustrates
that the number of spanning trees of the kite graph
is f (1, 1) = 8 as you see the 8 possible trees.
Solution:
The gradient of f is ∇f = h1 + 4x + 3x2 + 2y, 1 + 2x + 2yi.
Setting the gradient equal to ~0 and solving for the critical points,
we get: (0, − 21 ) and(− 32 , 16 ).
To classify the critical points, we compute the matrix of second
derivatives, which is
!
4 + 6x 2
2
2
At (0, − 21 ), we get
4 2
2 2
!
which has a positive determinant and fxx > 0, so the point
(0, − 12 ) is a local minimum.
At (− 23 , 61 ), we get
!
0 2
2 2
which has a negative determinant, so the point (− 32 , 16 ) is a
saddle point.
Main definitions
Standard assumption is always that functions are
smooth in the sense that all first and second partial
derivatives are continuous.
A point (x0, y0) is a critical point of f if
∇f (x0, y0) = h0, 0i.
Fermat’s theorem: if f has a local maximum or
local minimum at (x0, y0) then (x0, y0) is a critical
point
Second derivative test: Assume (x0, y0) is a critical point. If D < 0 then it is a saddle point. If
D > 0, fxx < 0 then (x0, y0) is a local maximum. If
D > 0, fxx > 0 then (x0, y0) is a local minimum.

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