Assignment 5
Math 110-003
Due Thurs Mar 31
Hand in full solutions to the questions below. Make sure you justify all your work and include
complete arguments and explanations. Your answers must be clear and neatly written, as well
as legible (no tiny drawings or micro-handwriting please!). Your answers must be stapled,
with your name and student number at the top of each page.
1. This problem is a variant of the “silly walks” problem from the workshop.
Suppose you are in a room that is a × 10 meters. You can “wall walk” along the a
meter long wall at s meters per second and floor walk in any direction at 1 meter
per second. You start at one corner of the room, wall walk along the wall of length
a meters for some distance, and then floor walk straight to the corner opposite from
where you started.
Suppose that the fastest route to the opposite corner of the room is to directly floor
walk from one corner of the room to the other (i.e. you wall walk for 0 distance).
What can be said about the relationship between a and s?
2. You can solve multivariable optimization problems by first treating one of the variables
as a parameter. This problem will illustrate how this is done.
We want to solve the following optimization problem: suppose x, y, and z are positive
numbers that sum to 10. What is the largest possible value of xy + xz + yz?
To solve this optimization problem, use the following steps:
(a) First, suppose that z is a fixed parameter. Then we have to find nonnegative
numbers x and y (depending on the fixed value z such that x + y = 10 − z), such
that xy + xz + yz is maximized.
(b) To find the global maximum, plug in the values for x and y obtained in (a) in
order to rewrite xy + xz + yz as a function of a single variable. Then maximize
this function.
Solution.
(a) First, use the constraint x + y = 10 − z to write y in terms of x (one could also
write x in terms of y):
y = 10 − z − x.
(1)
1
Plug this expression for y into the function xy + xz + yz (which is a function of
two variables x, y with one parameter z) to get the function
f (x) = x(10 − z − x) + xz + (10 − z − x)z,
(2)
which is a function of one variable (with one parameter z). We can expand,
simplify, and group like terms to get
f (x) = −x2 + (10 − z)x + (10z − z 2 ).
(3)
We wish to locate x (in terms of z) that maximizes this function, so we differentiate
it (with respect to x) and solve for the critical points. Since z is a constant with
respect to x,
f 0 (x) = −2x + (10 − z).
(4)
Thus, f 0 (x) = 0 when
−(10 − z)
1
= (10 − z).
(5)
−2
2
Since f 00 (x) = −2 < 0 for any x, this is the location of the global maximum.
Plugging this value for x back into the expression for y in terms of x, we get the
corresponding y value of
x=
1
1
y = 10 − z − x = (10 − z) − (10 − z) = (10 − z).
2
2
We conclude that
1
x = y = (10 − z).
2
(6)
(7)
(b) Plugging the values
1
1
x = y = (10 − z) = 5 − z
2
2
found above into xy + xz + yz, we get the function
xy + xz + yz = x2 + 2xz
= x(x + 2z)
= (5 − z/2)(5 − z/2 + 2z)
= (5 − z/2)(5 + 3z/2)
3
= − z 2 + 5z + 25.
4
(8)
(9)
(10)
(11)
(12)
(13)
Let g(z) be this function of z, i.e.
3
g(z) = − z 2 + 5z + 25.
4
(14)
We wish to maximize g(z), so we find the critical points. Solving
3
g 0 (z) = − z + 5 = 0,
2
2
(15)
we get
z=
10
.
3
(16)
3
Since g 00 (z) = − < 0 for all z, this is the location of the global maximum.
2
Plugging this value of z back into the expression for x and y in terms of z, we get
the corresponding values
1
1
10
1
x = y = (10 − z) = (10 − 10/3) = (20/3) = .
2
2
2
3
We conclude that
x=y=z=
3
10
.
3
(17)
(18)