Optimization
1. Find the point on the plane −3x + 2y + 3z = 1 which is closest to the point 4, −5, −2
Answer:
Minimize x − 4 2 + y + 5 2 + z + 2 2 subject to the constraint −3x + 2y + 3z − 1 = 0. Using
the cross product trick,
x − 4, y + 5, z + 2 × −3, 2, 3 =
3y + 11 − 2z, −3z + 6 − 3x, 2x + 7 + 3y = 0, 0, 0. Hence the point of interest
2
is 11
, − 27
, 20 .
11 11
2. Find the point of y 2 x = 16 which is closest to 0, 0.
Answer:
minimize x 2 + y 2 subject to xy 2 − 16 = 0. Then we need
x, y, 0 × y 2 , 2xy, 0 = 0, 0, 2x 2 y − y 3  = 0, 0, 0.
Hence we need y 2 = 2x 2 and so x = 2 from the constraint equation. Then the point is 2, 2 2
another point is 2, −2 2
3. Find three numbers x, y, z such that x 2 + y 2 + z 2 = 1 and 3x + 2y − 3z is as large as possible.
Answer:
Using the cross product trick, we need x, y, z × 3, 2, −3 =
−3y − 2z, 3z + 3x, 2x − 3y = 0.
Solution is : x = −z, y = − 23 z, z = z
satisfied and so
z 2 22
= 1 and so
9
z=±
1
3
22
22 ,
22
9
Then we also need to have the constraint equation
. Thus there are two points to check. − 223 22 , − 111 22 ,
1
11
3
22
22 , and
3
22 , − 22
22 . You just have to check to see which works better.
4. Find the points where possible local minima or local maxima occur in the function fx, y =
2yx + 6x + 12y + 2x 2 − 4y 2
Answer:
y = 1, x = −2
5. Find three positive numbers x, y, z such that their product is as large as possible and
4x + 4y + 2z = 5.
Answer:
Maximize xyz subject to 4x + 4y + 2z = 5. Recall how you can use the cross product to get
started on one of these.
yz, zx, xy × 4, 4, 2 = 2xz − 4yx, 4yx − 2yz, 4yz − 4xz
2xz − 4yx = 0
4yx − 2yz = 0
and so we need
4yz − 4xz = 0
, The interesting solution is z =
5
6
,y =
5
12
,x =
5
12
. Therefore,
4x + 4y + 2z = 5
the maximum value is
125
864
.
6. Minimize 16x 2 − 8yx + y 2 subject to the two constraints. 3x + 5y + z = 5 and −3x + y + z = 3.
Does the function have a maximum?
Answer:
The sensible thing to do in this example is to solve the constraints parametrically. Thus you need
3x + 5y + z = 5
, Solution is : y = − 13 z + 43 , x = 29 z − 59 Now plug this in. You want to
−3x + y + z = 3
2
minimize 11
z − 32
. This is now a function of a single variable. You just take the derivative
9
9
and set equal to 0. This yields
11
2 11
z − 32
=0
9
9
9
and this requires that z =
1
11
,
4
11
,
32
11
. Therefore, the minimum of this function occurs at
32
11
The value of this minimum is 0.
It has no maximum.
7. Find the points where possible local minima or local maxima occur in the function
fx, y = 7x 4 − 56x 2 + 112 + 20yx 2 − 80y + 16y 2
Next classify these points determining whether they are local minima, local maxima or saddle
points using the second derivative test.
Answer:
The gradient is
28x 3 − 112x + 40yx, 20x 2 − 80 + 32y
0,
5
2
, 2, 0 , −2, 0 are the critical points. Then the Hessian matrix is
84x 2 − 112 + 24y 40x
40x
At the point 0,
−52
0
0
32
32
5
2
this matrix equals
Now simply look at the two eigenvalues which are the diagonal entries of the above matrix and
use the rule. If both are positive, then it is a local minimum. If both are negative, then it is a local
maximum and if they have different signs, then it is a saddle point. Next consider the other points.
At 2, 0 the Hessian matrix is
224 80
80
32
while at −2, 0 it is
224 −80
−80
32
In both cases the determinant is 768. If this is negative, then the two eigenvalues have opposite
signs and so we have a saddle point. If this determinant is positive, then consider the trace, 256. If
it is positive, then the point is a local minimum and if negative, then the point is a local maximum.
8. Find the point on y = 4x + 6 which is closest to the point 1, 9
Answer:
2
Minimize x − 1 2 + y − 9 subject to the constraint 4x + 5 − y = 0. Using the cross product
trick,
2x − 1, 2 y − 9 , 0 × 4, −1, 0 = 0, 0, −2x + 74 − 8y
and so we need the two equations −2x + 74 − 8y = 0 and y = 4x + 6. The solution is
x = 13
, y = 154
.
17
17
9. The function fx, y = −2x 2 + 3y 2 is defined on the disk x 2 + y 2 ≤ 1. Find the minimum value of
this function on this disk and explain why it has a minimum value.
Answer:
∇ −2x 2 + 3y 2 = −4x, 6y which is only equal to 0 at 0, 0. Therefore, this point is of no
interest because it is clear that we can find points where the function is smaller than 0. It follows
we need to minimize −2x 2 + 3y 2 subject to the constraint x 2 + y 2 = 1. Using the cross product
trick,
−4x, 6y, 0 × 2x, 2x, 0 = 0, 0, −8x 2 − 12yx = 0
And so we need to have −2x 2 − 3yx = 0 which happens if x = 0 and y = ± 13 3 or if x ≠ 0 and
−2x = 3y. In this case, the constraint equation requires
2
3
y = ±− 13
13 , x = ± 13
13
Thus the list of candidates is 0, 1, 0, −1,
3
− 13
13 ,
2
13
3
13
13 , − 132 13 , and
13
Now it is just a matter of plugging these in and seeing which one gives the smallest value.
10. Find the area of the largest rectangle which can be inscribed in the ellipse 161 x 2 + 19 y 2 = 1.
Answer:
The area of this rectangle is 2x2y where x, y is the point obtained from maximizing xy
1 2
subject to the constraint 16
x + 19 y 2 − 1 = 0. Thus
y − λ2x/a = 0, x − λ2y/b = 0.
Thus, assuming λ ≠ 0 which must happen after all, in order to satisfy the constraint, we must
2
2
2
x2
have xa = yb and so from the constraint equation, ax 2 + ab
= 1, so x = 47 21 , y = 67 7 .
Therefore, the maximum area is
96
7
3.
11. Find the volume of the largest box which can be inscribed in the ellipsoid
1
16
x2 +
1
4
y 2 + z 2 = 1.
Answer:
The idea is to maximize xyz subject to the constraint 161 x 2 + 14 y 2 + z 2 = 1 and then the volume of
this box will be 8 times that number. Thus
1 2
L = xyz − λ 16
x + 14 y 2 + z 2 − 1 and so we need to solve
yz −
1
8
λx = 0
xz −
1
2
λy = 0
yx − 2λz = 0
1
16
x2 +
1
4
y2 + z2 − 1 = 0
Multiply the top by x, the second by y, and the third by z. Thus
1
16
x2 =
y=
2
3
1
4
y 2 = z 2 . Hence 3
3 and z =
1
3
1
16
x2
= 1 and so in the first quadrant, x =
3 . Therefore, the maximum volume is
64
9
4
3
3 and similarly
3.
12. Two lines have parametric equations 1, 0, 0 + t 2, −3, 1 and 0, −3, −2 + t 0, 1, 1 . Find
points, one from the first and one from the second which are as close as possible.
Answer:
You shouldn’t use Lagrange multipliers on this one. You want to minimize
ft, s = 1 + 2t 2 + 3 − 3t − s 2 + 2 + t − s 2
Take the gradient and set equal to 0. Then solve for the correct values of t, s. The gradient is
−10 + 4t + 4s, −10 + 28t + 4s
−10 + 4t + 4s = 0
−10 + 28t + 4s = 0
, The solution is
t = 0, s =
5
2
The points are 1, 0, 0
0, − 12 ,
1
2
13. Find a point on the level surface 2x 2 + 2y 2 + 3z 2 = 1 which is closest to the point 0, 0, 0.
Answer:
Minimize x 2 + y 2 + z 2 subject to the constraint 2x 2 + 2y 2 + 3z 2 − 1 = 0. Using the cross product
trick,
x, y, z × 2x, 2y, 3z = yz, −zx, 0 = 0, 0, 0.
The first component requires yz = 0. If both y, z = 0 then we need to look at ±
1
2
, 0, 0 if it
makes sense. Suppose then that z ≠ 0 but y = 0. Then from the second component, x = 0 and we
must look at 0, 0, ±
1
3
if it makes sense. Finally, suppose z = 0, y ≠ 0. Then we need to have
2x 2 + 2y 2 = 1 and so many points of the form
1
2
Here t ∈ 0, 2π. For the first case, the distance is
this makes sense. In the third, it is
1
2
2 cos t,
1
2
1
2
2 sin t, 0 would be considered.
2 , in the second case, it is
1
3
provided
. You pick a point which gives the smallest answer.
14. A point in the first quadrant is picked on the ellipse 4x 2 + y 2 = 1. From this point, the tangent line
to the ellipse is drawn. This line forms a triangle when it is included with the x and y axis. Find the
area of the smallest such triangle.
Answer:
The equation of the line can be found using implicit differentiation. For a point x, y on this curve,
2
2
the slope of the tangent line at this point is −4 xy . The line intersects the x axis at the point 14 4x x+y
and the y axis at the point
4x 2 +y 2
y
. Therefore, we want to maximize
1
8
4x 2 +y 2
xy
2
subject to the
constraint 4x 2 + y 2 = 1.
1
8
12x −y
4x 2 + y 2  x 2 y − 8λx = 0
1
8
3y −4x
4x 2 + y 2  xy 2 − 2λy = 0
2
2
2
2
Multiply the top equation by
1 3y 2 −4x 2
4
xy 2
and the bottom by
1 12x 2 −y 2
4
x2y
. Then the messy stuff on the
left in both of these is the same and if follows that
2λ 3y y−4x
=
2
2
2
3y 2 −4x 2
y2
=
1
2
λ 12xx 2−y
2
1 12x 2 −y 2
4
x2
2
, so y = 2x .
Next put this in to the constraint equation to find x.
8x 2 = 1, so x =
1
4
2 and y =
1
2
The smallest such triangle is then
2
1
2
.
15. Find the volume of the largest circular cylinder which can be inscribed in the ball x 2 + y 2 + z 2 ≤ 4
Answer:
Maximize 2πr 2 z subject to the constraint r 2 + z 2 = 4. Of course it suffices to only consider r 2 z
since this would be simpler. Thus,
2rz, r 2 , 0 × 2r, 2z, 0 = 0, 0, 4rz 2 − 2r 3  = 0, 0, 0
We know r ≠ 0. Therefore, 4z 2 = 2r 2 , so z = 12 2 r. Then from the constraint equation,
r2 +
32
9
1
2
2
2r
= 4, so r =
6 and hence z =
2
3
2
3
3 . It follows that the maximum volume is
π 3.
16. A cylindrical can is supposed to have a volume of 49π cubic centimeters. The top and bottom are
made of a material which costs 2 cents per square centimeter and the sides are made of a material
which costs 1 cents per square centimeter. Find the dimensions of the cheapest possible can
having the given volume.
Answer:
Minimize 2πry + 4πr 2 subject to the constraint πr 2 y = 49π. Then using the trick with the
gradient, we need
y + 4r, r, 0 × 2ry, r 2 , 0 = 0, 0, −r 2 y + 4r 3 = 0, 0, 0
and so we need 4r = y. Now from the constraint, we need r 2 4r
r=
1
2
3
2
3
7
2
,y = 2 3 2
3
7
2
= 49, so
17. Maximize −x − 4y subject to the constraint
Answer:
−1, −4, 0 ×
1
4
x, 14 y, 0
1
4
x2 +
1
4
y 2 = 1.
= 0, 0, − 14 y + x
Hence − 14 y + x = 0, Solution is : y = 4x . Then from the constraint equation,
17
4
2
x 2 = 1, Solution is : x = − 17
17 , x =
2
8
− 17
17 , − 17
17 ,
2
17
17 ,
8
17
17
2
17
17 and so the points of interest are
You plug these in and see which one works best. One
will give the maximum and other will give the minimum.
18. Find the points where possible local minima or local maxima occur in the function
fx, y = −x 4 + 8x 2 − 16 − 2y 2
Next classify these points, determining whether they are local minima, local maxima or saddle
points, using the second derivative test.
Answer:
The gradient is
−4x 3 + 16x, 4y
0, 0 , 2, 0 , −2, 0 are the critical points. Then the Hessian matrix is
−12x 2 + 16 + 8y 0
0
4
At the point 0, 0 this matrix equals
16 0
0
4
Now simply look at the two eigenvalues which are the diagonal entries of the above matrix and
use the rule. If both are positive, then it is a local minimum. If both are negative, then it is a local
maximum and if they have different signs, then it is a saddle point. Next consider the other points.
At 2, 0 the Hessian matrix is
−32 0
0
4
while at −2, 0 it is
−32 0
0
4
In both cases the determinant is −128 If this is negative, then the two eigenvalues have opposite
signs and so we have a saddle point. If this determinant is positive, then consider the trace, −28. If
it is positive, then the point is a local minimum and if negative, then the point is a local maximum.
19. A cylindrical can is supposed to have a volume of 49π cubic centimeters. Find the dimensions of
the can which minimizes surface area.
Answer:
Minimize 2πry + 2πr 2 subject to the constraint πr 2 y = 49π. The Lagrangian is then
ry + r 2 − λr 2 y + 49λ. Then using the trick with the gradient, we need
y + 2r, r, 0 × 2ry, r 2 , 0 = 0, 0, y + 2rr 2 − 2r 2 y = 0, 0, 0. Hence we need
2r = y. Now from the constraint, we need r 2 2r = 49, and so r =
y=
3
2
2
3
2
7
20. Find the point on
1
4
1
2
3
2
2
3
7
2
and
.
x2 +
1
9
y2 +
1
4
z 2 = 1 which is closest to the plane
1
4
x+
1
9
y+
1
10
z = 1.
Answer:
First of all, if x 0 , y 0 , z 0  is a point, how close is it to the plane? The square of its distance is
1
− 90x 0 − 40y 0 − 36z 0  2 . Therefore, we want to minimize
10 996 360
1
10 996
360 − 90x − 40y − 36z 2 subject to the constraint
need
200
− 162749
+
4050
2749
x+
y+
1800
2749
1620
2749
1
4
x2 +
1
9
y2 +
1
4
z 2 − 1 = 0. Thus we
z − λx = 0
− 7200
+
2749
1800
2749
x+
800
2749
y+
720
2749
z−
− 6480
+
2749
1620
2749
x+
720
2749
y+
648
2749
z − λz = 0
2
3
λy = 0
Divide the top equation by 90, the next by 40, and the last by 36. Then assuming λ ≠ 0, this
leads to
1
1
1
x = 120
y = 72
z
180
and so y =
2749
8100
2
3
x, z =
x 2 = 1, x =
90
2749
2749 ,
2
5
90
2749
60
2749
and −1 times this.
x. Now from the constraint equation,
90
2749 , x = − 2749
2749 Hence the two points of interest are
2749 ,
36
2749
2749