Problem set 2 due Friday June 1 by 7 pm in class or in the mailbox.
Problem 1.
Compute the determinants of the following matrices.
(a)


1 0 1
 1 0 1 
0 1 1
(b)


1 2 0
 1 2 3 
1 0 4
(c)


−6 −3 0
 3
1 0 
0
0 3
(d)


2 −1 −3
 1
4 −1 
−1 2
0
(e)


1 −1 −3
 2 0
4 
1 0
2
(f)


−2 0 0
 0
2 1 
0 −1 2
Hint : Two of those determinants can be directly seen to be 0 because of linear dependence.
Two of these matrices are actually block diagonal, and thus you can reduce to computing the determinant of a 2 × 2 matrix. For the other two determinants, you can either use the general formula, or
expand the determinant along a well chosen line or column.
Problem 2.
Let f (x, y) be a nicely behaved function of two variables of which you only know that f (1, 0) = 2,
fx (1, 0) = 0 and fy (1, 0) = 1.
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(a) What’s your best guess for f (1.01, −0.03) ?
(b) You are now told that fxx (1, 0) = −1, fxy (1, 0) = 1 and fyy (1, 0) = 1. What is your answer to
the previous question now ?
Problem 3.
Compute the Taylor approximation of order two of the following functions.
(a)
sin(x)
ex
at x = 0
(b) ln(x + ey ) at (x, y) = (0, 0)
(c) xy at (x, y) = (1, 1)
Problem 4.
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Consider the function f (x) = x3 + 2x2 defined on the closed interval [−5, 5].
Which interior points are local minima ? Which ones are local maxima ?
What are the global extrema and what are the minimum and maximum values of f on [−5, 5] ?
Hint : Don’t forget the two boundary points.
Problem 5. On Fermat’s principle.
You’re a lifeguard on the beach at point A of coordinates (0, h1 ), and spot someone seemingly in
trouble in the sea at point B = (l, −h2 ). You need to minimize you travel time to this point. The
beach/sea interface is the horizontal line y = 0. You’ll run in a straight line to a point P = (x, 0) and
from there swim in a straight line to point B.
(a) Express d1 , the distance between A and P as a function of the parameters introduced above.
Compute similarly d2 , the distance between P and B
Hint : You can use Pythagoras’ theorem.
(b) Recall that the time of travel is given by length = speed × time. Write down the time t1 it takes
you to cover the distance AP on the beach, in terms of d1 and of your running speed, s1 . Compute
similarly the time t2 it takes you to cover the distance P B in the sea, knowing your swimming speed is s2 .
(c) As a lifeguard, you’re required to minimize your total time of travel T (x) = t1 + t2 by chosing in
a clever way the best value for x, i.e. the best position for the point P where you’re going to enter the
sea.
Consider the incidence angles θ1 and θ2 (in other words, θ1 (respectively θ2 ) is the angle the segment
AP (respectively BP ) makes with a vertical line). Check, using trigonometry in relevant triangles, that
sin θ1 = x/d1 , and sin θ2 = (l − x)/d2 .
Explain using a calculus computation why Snell’s law (s2 sin θ1 = s1 sin θ2 ) is a necessary condition
on x for the total time function T (x) to have a local extrema at x.
Hint : if you take a derivative, don’t forget that quantities such as d1 or θ1 are functions of x.
(d) We now want to prove that T has a unique global minimum.
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If x < 0 or x > l, show - by a quick common sense argument - that T attains values strictly smaller
than T (x) on the interval [0, l].
Show that on [0, l], sin θ1 is increasing in x whereas sin θ2 is decreasing in x. Use this to show that
there is a unique point x∗ in [0, l] for which Snell’s law is satisfied (you can for example consider the
quantity sin θ1 / sin θ2 ).
(e*) Show that T has a unique global minimum, at x∗ .
You can for example use the results of (d) and try to adapt to the function T (x) the "compactness"
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argument given in class for the function of two variables ex + y 2 .
Problem 6. A naive approach to a minimization problem with constraint.
Consider a rectangular parallelepiped i.e. a box, the faces of which are rectangles. Let us call its
edge sizes x, y and z.
(a) Express the surface area A (the box is closed : it has six faces), and the volume V of the box as
functions of the three edge sizes.
(b) We want to minimize the surface area A(x, y, z), for a fixed volume V (x, y, z) = V0 (i.e. we
restrict the domain of A to those points (x, y, z) satisfying the equation V (x, y, z) = V0 ).
Under the constraint V = V0 , express z as a function of x, y and V0 , and use this to rewrite A as a
function of the two variables x and y.
Show that A(x, y) admits a unique local minimum.
(c**) Discuss globality. You can try a "compactness" argument i.e. mimick what I did in class for
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the function ex + y 2 . Recall that the core of the argument was to show that out of a region not too big,
the function takes values too big to matter when looking for a minimum. For example, you can :
- Assume c ≥ x > 0. Check A ≥ V /c.
- Show (by a drawing in the xy plane) that the three conditions x ≥ c, y ≥ c and z ≥ c cut out a
region fitting our purpose (i.e. a region that stays away from infinity, and from the boundaries x = 0
and y = 0).
-Conclude.
(d) We would now like to solve a very related problem, namely maximizing the volume V (x, y, z),
for a fixed surface area A(x, y, z) = A0 .
Express z as a function of x, y and A0 . Notice that it is not a very friendly-looking formula.
One can still solve the problem with the technique used in (b) but the computation is quite messy.
We’ll thus use a somewhat indirect approach (we’ll see later in class a technique to solve this problem
with a direct but much easier computation).
(e) Using the result of (b), we’re going to prove that x = y = z =
with constraint A = A0 .
p
A0 /6 is a global maximum of V
0 0 0
0 0 0
Argue by contradition
p
p: suppose
p you can find a point (x , y , z ) satisfying A(x , y , z ) = A0 and
0
0
0
V (x , y , z ) > V ( A0 /6, A0 /6, A0 /6).
By rescaling the box (x0 , y 0 , z 0 ), find a contradiction with the result of (b).
Conclude.
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