Download Homework 4 Solutions

Homework 4 Solutions
May 6, 2017
Problem 1
a)
Let f (x, y) = x − y and g(x, y) = x2 − y 2 .
We need to find the extrema of f (x, y) subject to the constraint g(x, y) = 2.
We first use the method of Lagrange multipliers to find all critical points (potential extrema).
Thus, we need to find (x, y) ∈ R2 and λ ∈ R such that the following equations hold:
∇f (x, y) = λ∇g(x, y)
g(x, y) = 2
Writing this out, we obtain the system:
1 = λ2x
−1 = −λ2y
x2 − y 2 = 2
The first two equations imply that x = y, which is incompatible with the third equation.
So the system has no solutions. This implies there are no critical points and so f has no
extrema subject to the given constraints.
b)
Let f (x, y, z) = x + y + z, g(x, y, z) = x2 − y 2 , and h(x, y, z) = 2x + z.
We need to find the extrema of f (x, y, z) subject to the constraints g(x, y, z) = 1 and
h(x, y, z) = 1.
We first use the method of Lagrange multipliers to find all critical points.
Thus, we need to find (x, y, z) ∈ R3 and λ1 , λ2 ∈ R such that the following equations
hold:
∇f (x, y) = λ1 ∇g(x, y) + λ2 ∇h(x, y)
g(x, y) = 1
h(x, y) = 1
Writing this out, we obtain the system:
1 = λ1 2x + 2λ2
1 = −λ1 2y
1 = λ2
x2 − y 2 = 1
2x + z = 1
1
The first 3 equations together imply that x = y, which is incompatible with the 4th
equation. So the system has no solutions. This implies there are no critical points and so
f has no extrema subject to the given constraints.
Problem 2
Let f (x, y) = x2 + y 2 − x − y + 1 and D = {(x, y) ∈ R2 : x2 + y 2 ≤ 1}.
We need to find the absolute minimum and maximum of f on D.
First we find all critical points of f on D.
For points in the interior of D we use the first derivative test and find (x, y) ∈ interior(D)
such that:
∇f (x, y) = ~0
Writing this out, we obtain the system:
2x − 1 = 0
2y − 1 = 0
The unique solution to this system is x = y = 21 , so the only critical point in the interior
is ( 21 , 12 ).
For the critical points on the boundary of D we use the method of Lagrange multipliers
subject to the constraint g(x, y) = 1, where g(x, y) = x2 + y 2 .
Thus, we need to find (x, y) ∈ R2 and λ ∈ R such that the following equations hold:
∇f (x, y) = λ∇g(x, y)
g(x, y) = 1
Writing this out, we obtain the system:
2x − 1 = λ2x
2y − 1 = λ2y
x2 + y 2 = 1
The first two equations imply that x = y, and plugging this in the 3rd equation we see
that the system has two solutions: x = y = √12 and x = y = − √12 .
So the the two critical points on the boundary of D are ( √12 , √12 ) and (− √12 , − √12 ).
In total there are 3 critical points (potential extrema) on D: ( 12 , 21 ), ( √12 , √12 ), and (− √12 , − √12 ).
Next we note the D is closed and bounded, and f is continuous, so absolute extremum
of f on D must exist. To find them we need only evaluate f at all the critical points and
take the largest/smallest value.
1 1
1
f( , ) =
2 2
2
√
1 1
f(√ , √ ) = 2 − 2
2 2
√
1
1
f (− √ , − √ ) = 2 + 2
2
2
2
Thus on D, f obtains an absolute minimum of
√
2 + 2 at (− √12 , − √12 ).
1
2
at ( 12 , 21 ) and an absolute maximum of
Problem 3
Denote by w, h, and d be the width, height, and depth (respectively) of the box. Thus
the surface area is A(w, h, d) = 2wh + 2hd + wd and the volume is V (w, h, d) = whd.
We want to find the maximum of V (w, h, d) subject to the constraint A(w, h, d) = 16.
We first use the method of Lagrange multipliers to find all potential such maxima.
Thus, we need to find (w, h, d) ∈ R3 and λ ∈ R such that the following equations hold:
∇V (w, h, d) = λ∇A(w, h, d)
A(w, h, d) = 16
Writing this out, we obtain the system:
wd = λ(2w + 2d)
hd = λ(2h + d)
hw = λ(2h + w)
2hw + 2hd + wd = 16
Rearranging the first 3 equations so that only λ1 is on the left, and then setting these equal
to each other, we get 2h = w = d. Plugging this into the 4th equation gives 3w2 = 16.
Hence the only solution to the system is (w, h, d) = ( √43 , √23 , √43 ), and this is our critical
point.
The last thing to do is justify why this point must indeed be a maximum. This is done
by applying theorem 10 on page 198, or more precisely it’s generalization on page 199.
Basically, one needs to write down the bordered Hessian at ( √43 , √23 , √43 ), compute the
determinants of the relevant sub matrices, and verify that the conditions for a maximum
are satisfied. This a tedious but straightforward computation that can be done using a
computer.
(Another way to do this is to note that we have some additional constraints, namely
w ≥ 0, h ≥ 0, and d ≥ 0. This is because lengths must be positive. Combined with the
, V ≤ 64
and V ≤ 64
. So V
constraints given by A this implys the estimates V ≤ 64
w
h
d
subject to all these constraints is non-negatve, continuous, and vanishes at infintiy, which
implies a global maximum must exist.)
In conclusion, the dimensions that maximize the volume are (w, h, d) = ( √43 , √23 , √43 ), and
√ .
the maximum volume is 332
3
Problem 4
Let F (x, y, z) = x3 z 2 − z 3 yx, and consider the equation F (x, y, z) = 0.
First we attempt to solve for z in terms of (x, y) near the point (1, 1, 1). We want to apply
the Implicit Function Theorem, so we need to compute ∂F
(1, 1, 1):
∂z
∂F
3
2
(1, 1, 1) = 2x z − 3z yx
= −1
∂z
(x,y,z)=(1,1,1)
Since this is non-zero, we conclude that yes, we can solve for z in terms of x and y near
(1, 1, 1).
∂z
∂z
Next we compute ∂x
and ∂y
at (1, 1).
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From our work above, we know that near (1, 1) we can write z = z(x, y), and that
F (x, y, z(x, y)) = 0. Note also that z(1, 1) = 1. Thus we have the following equation near
(1, 1):
x3 z(x, y)2 − z(x, y)3 yx = 0
Differentiating both sides of this with respect to x and y gives:
∂z
∂z
(x, y) − 3z(x, y)2 (x, y)yx − z(x, y)3 y = 0
∂x
∂x
∂z
∂z
x3 2z(x, y) (x, y) − 3z(x, y)2 (x, y)yx − z(x, y)3 x = 0
∂y
∂y
3x2 z(x, y)2 + x3 2z(x, y)
Evaluating at (x, y) = (1, 1) and using the fact that z(1, 1) = 1, we obtain:
∂z
∂z
(1, 1) − 3 (1, 1) − 1 = 0
∂x
∂x
∂z
∂z
2 (1, 1) − 3 (1, 1) − 1 = 0
∂y
∂y
3+2
Rearranging we obtain:
∂z
(1, 1) = 2
∂x
∂z
(1, 1) = −1
∂y
Finally we show that we cannot solve for z in terms of (x, y) near (0, 0, 0). To see this
we simply note that (0, 0, t) is a solution to our equation for every t ∈ R. Thus the set of
solutions to our equation contains a vertical line through the point (0, 0, 0), so near this
point it fails the vertical line test. Hence, it’s impossible to solve for z in terms of (x, y)
near (0, 0, 0).
Problem 5
Consider the function F : R5 → R2 , F (x, y, z, u, v) = (xy 2 + xzu + yv 2 , u3 yz + 2xv − u2 v 2 ),
and also the equation F (x, y, z, u, v) = (3, 2).
We would like to solve this equation for (u, v) in terms of (x, y, z) near the point (1, 1, 1, 1, 1).
We want to apply the Implicit Function Theorem, so we need to evaluate the determinant
of the following matrix at (1, 1, 1, 1, 1):
xz
2yv
3u2 yz − 2uv 2 2x − 2u2 v
Evaluating at (1, 1, 1, 1, 1) we get:
1 2
1 0
The determinant of this matrix is −2 which is non-zero. Thus we conclude that yes, it is
possible to solve for (u, v) in terms of (x, y, z) near the point (1, 1, 1, 1, 1).
∂v
at (1, 1, 1).
Next we compute ∂y
From our work above, we know that near (1, 1, 1) we can write (u, v) = (u(x, y, z), v(x, y, z)),
4
and that F (x, y, u(x, y, z), v(x, y, z)) = (3, 2). Note also that (u(1, 1, 1), v(1, 1, 1)) = (1, 1).
Thus we have the following system of equations near (1, 1, 1):
xy 2 + xzu(x, y, z) + yv(x, y, z)2 = 3
u(x, y, z)3 yz + 2xv(x, y, z) − u(x, y, z)2 v(x, y, z)2 = 2
Differentiating both equations with respect to y we obtain:
∂u
∂v
(x, y, z) + 2yv(x, y, z) (x, y, z) + v(x, y, z)2 = 0
∂y
∂y
∂u
∂v
u(x, y, z)3 z + 3u(x, y, z)2 (x, y, z)yz + 2x (x, y, z)
∂y
∂y
∂u
∂v
−2u(x, y, z) (x, y, z)v(x, y, z)2 − 2v(x, y, z) (x, y, z)u(x, y, z)2 = 0
∂y
∂y
2xy + xz
Evaluating at the point (x, y, z) = (1, 1, 1) and using the fact that (u(1, 1, 1), v(1, 1, 1)) =
(1, 1) we obtain the following system of equations:
∂u
∂v
(1, 1, 1) + 2 (1, 1, 1) + 1 = 0
∂y
∂y
∂u
∂v
∂u
∂v
1 + 3 (1, 1, 1) + 2 (1, 1, 1) − 2 (1, 1, 1) − 2 (1, 1, 1) = 0
∂y
∂y
∂y
∂y
2+
Solving the system we obtain
∂v
(1, 1, 1)
∂y
= −1
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