Covers: 13.8 and 13.10
Grade 1(a), 1(b), 3, 8, 9 at 10 points each
Exercise 1. Test the function for relative extrema and saddle points. (a)f(x, y) = 2x4+y2−x2−2y
(b)f(x, y) =x3−3xy+y2+y−5 (c)f(x, y) = 4xy−x4−y4
(d)x+y− 1
xy
Solution 1. (a)
fx= 8x3−2x= 0 at x= 0 and x=±1/2 fy = 2y−2 at y= 1
So the critical points are (0,1), (1/2,1), and (−1/2,1).
fxx = 24x2−2 fyy = 2 fxy =fyx= 0 sod(x, y) =fxxfyy−(fxy)2= 48x2−4.
d(0,1) =−2 so a saddle occurs at (0,1). d(±1/2,1) = 8 andfyy >0 so local minima occur at (±1/2,1).
(b) Solve
fx= 3x2−3y= 0 fy =−3x+ 2y+ 1 = 0
Soy= 3/2x−1/2 and hence 3x2−9/2x+ 3/2 = 3/2(2x2−3x+ 1) = 3/2(2x−1)(x−1). Sox= 1,1/2 and we have the critical points (1,1) and (1/2,1/4).
fxx = 6x fyy = 2 fxy =fyx=−3
(c) Solve
fx = 4y−4x3 = 0 fy = 4x−4y3 = 0
So y = x3 and hence x = y3 = x9. So x = 0 or x = ±1. This gives rise to the critical points (0,0), (1,1), and (−1,−1).
Next check the nature of these critical points:
fxx=−12x2 fyy =−12y2 fxy =fyx= 4
and d(x, y) = 122x2y2−4. Sod(0,0) = −4 and (0,0) is a saddle; d(−1,−1) = d(1,1) = 140>0 and fxx(−1,−1) =fxx(1,1) =−12 so (−1,−1) and (1,1) are local maxima.
(d) Solve
fx= 1 + 1 x2y = 0
fy = 1 + 1 xy2 = 0
We have
fxx=− 2
x3y −
2
xy3 fxy =fyx=
1 x2y2
We haved(x, y) = x44y4−x41y4 = x43y4 andd(−1,−1) = 3>0 so (−1,−1) andfxx(−1,−1) =
−2<0 so (−1,−1) is a local max.
Exercise 2. Find absolute extrema of f(x, y) = (x−y)e−x2−y2 on the triangular region with vertices (−1,−1), (1,−1), and (0,1).
Solution 2. First we find the local extrema fx=e−x
2−y2
+ (x−y)e−x2−y2(−2x) =e−x2−y2[1−2x(x−y)] =e−x2−y2[1−2x2+ 2xy] fy =−e−x
2−y2
+ (x−y)e−x2−y2(−2x) =e−x2−y2[1−2y(x−y)] =e−x2−y2[−1 + 2y2−2xy]
So when fx =fy = 0 we have fx = 1−2x2+ 2xy = 0 =−fy = 1−2y2+ 2xy sox2 =y2 andx=±y. Settingx=ygives, 1−2x2+ 2x2 = 0 which is impossible and settingx=−y gives 1−4x2 so x=±1/2. Thus the critical points are (1/2,−1/2) and (−1/2,1/2). The
critical point (1/2,−1/2) is inside the triangle.
To find the extrema along the boundary we must check the points (−1,−1), (1,−1), and (0,1) and the critical points along the boundary.
We need to parametrize the lines forming the sides of the triangle. I will parametrize these using y as a function ofx so that in each case I get
df
dx =fx+fy dy dx =e
−x2−y2h
1−2x2+ 2xy+ (−1 + 2y2−2xy)dydx i
so dxdf = 0 when
1−2x2+ 2xy= (1−2y2+ 2xy)dydx
The line connecting (0,1) and (1,−1) is y = −2x+ 1 for 0 ≤ x ≤ 1. For this we have df
dx = 0 when
Now substituting−2x+ 1 fory we get
3−2x2−4(−2x+ 1)2+ 6(x)(−2x+ 1) = 3−2x2−4(4x2−4x+ 1)−12x2+ 6x =−1 + 22x−30x2 = 0
The roots are
−22±p(−22)2−4(−1)(−30)
−60 =
11±p(7)(13)
30 ≈0.685,0.049
The line connecting (0,1) and (−1,−1) is y = 2x+ 1 for −1 ≤x ≤ 0. For this we have df
dx = 0 when
1−2x2+ 2xy= (1−2y2+ 2xy)(2) = 2−4y2+ 4xy
−1−2x2+ 4y2−2xy= 0 Now substituting 2x+ 1 fory we get
−1−2x2+ 4(2x+ 1)2−2x(2x+ 1) =−1−2x2+ 4(4x2+ 4x+ 1)−4x2−2x = 3 + 14x+ 10x2= 0
This has one root in [−1,0] atx=
√
19−7
10 =−0.26...
The line connecting (−1,1) and (1,−1) is y=−1 for−1≤x≤1. For this we have dxdf = 0 when
1−2x2+ 2xy= (1−2y2+ 2xy)(0) = 0 3−2x2+ 2xy= 0
Now substituting−1 fory we get
1−2x−2x2 = 0 This has roots
2±p
4−(4)(−2)(1)
−4 =
−1±√3 2 Of these −1±
√
3
Now you simply need to computef(x, y) at each of the relevant points:
x y f(x, y)
1/2 −1/2 0.607
11−√(7)(13)
15 −2
11−√(7)(13) 15
+ 1 0.575
11+√(7)(13)
15 −2
11+√(7)(13) 15
+ 1 -0.377
√
19−7 10
√
19−2
5 -0.55
−1−√3
2 −1 0.44
0 1 -0.369
−1 −1 0
1 −1 0.271
So the maximum occurs at (1/2,−1/2) and the minimum occurs along an edge at √
19−7 10 ,
√
19−2 5
.
Exercise 3. Find the maximum off(x, y) = 2xy subject to x2+ 2y2 = 4. Solution 3. Solve
2y=λ2x 2x=λ4y x2+ 2y2 = 4
These yield y=λx=λ(λ2y) = 2λ2y, so thatλ2= 1/2. It follows that y2 =λ2x2 = 1/2x2 and sox2+ 2y2 = 2x2 = 4, hence x2 = 2 and y2 = 1. This gives 4 points (√2,±1). Now just check
x y f(x, y)
√
2 1 2√2
√
2 −1 −2√2
−√2 1 −2√2
−√2 −1 2√2
Exercise 4. Use LaGrange multipliers to find all points on the ellipse 4x2+y2 = 8 where the productxy is a minimum.
Solution 4. Solve
y= 8λx x= 2λy 4x2+y2 = 8
Here y = 8λ(2λy) = 16λ2y, so that λ2 = 1/16. So x2 = 4λ2y2 = 1/4y2 and 4x2+y2 = 2y2 = 8. We havey=±2 and x=±1. this gives the points (±1,±2).
x y f(x, y)
1 2 2
1 −2 −2
−1 2 −2
−1 −2 2
The minimum is−2 and this occurs at (1,−2) and (−2,1).
Exercise 5. Minimizef(x, y, z) = 4x2+y2+ 5z2 subject to 2x+ 3y+ 4z= 12. Solution 5. Solution 1. (Using LM.) Solve
8x= 2λ 2y= 3λ 10z= 4λ 2x+ 3y+ 4z= 12 Soλ= 4x and y= 6x and z= 8/5x. So
2x+ 3y+ 4z= 2x+ 18x+ 32/5x= 132/5x= 12
Sox= 5/11,y= 30/11, andz= 8/11 is the only critical point. f(5/11,30/11,8/11) = 10 is the minimum value of f on 2x+ 3y+ 4z= 12. Why not a saddle point or a minimum? - I’ll leave that to the reader to ponder.
Solution 2. The constraint givesz= 3−1/2x−3/4y and so we can express f restricted to the constraint in terms ofx and y
g(x, y) =f(x, y,3−1/2x−3/4y) = 4x2+y2+ 5(3−1/2x−3/4y)2
Rather than working the algebra, I will use the chain rule - recalling that∂y/∂x=∂x/∂y= 0:
gx =fx+fz∂z
∂x = 8x+ 10z(−1/2) = 21
2 x+ 15
4 y−15 gy =fy+fz∂z
∂y = 2y+ 10z(−3/4) = 61
8 y+ 15
4 x− 45
Now find where gx = 0 = gy to be at x = 5/11 and y = 30/11. Now apply the second partials test tog
gxx= 21/2 gyy= 61/8 gxy =gyx= 15/4
d(x, y) = (21/2)(61/8)−(15/4)2 >0 and gxx >0 so (5/11,30/11) is the minimum for g and so (5/11,30/11,8/11) is the minimum forf on 2x+ 3y+ 4z= 12.
Exercise 6. Maximize f(x, y, z) = 32xyz subject to the constraints, x+y+z = 4 and
−x−y+z= 3.
Solution 6. This one does not require LaGrange. Adding together the two constraints gives 2z = 7 so z = 7/2 and so x = 1/2−y, hence f restricted to the two constraints can be expressed as a function of x, h(x) = f(x,1/2−x,7/2) = 32x(1/2−x)(7/2) = (7)(16)x(1/2−x) andh0(x) = (7)(16)[1/2−2x] and so h0(x) = 0 atx = 1/4. This is the unique local maximum of hand hence the maximum of f on the constraints.
Exercise 7. The temperature on the surface x2+y2+ 4z2 = 12 is given by T(x, y, z) =
x2+y2. Find the maximum temperature on this surface.
Solution 7. The constraint is justT+ 4z2 = 12 or equivalentlyT = 12−4z2 which clearly
has a maximum value of 12 on the circlex2+y2 = 12. If you use LaGrange this is exactly what you find.
Exercise 8. Use Lagrange multipliers to find three positive numbers whose sum is 33 and whose product is a maximum.
Solution 8. We wantxyz maximal whenx+y+z= 33. So solve yz=xz=xy =λ x+y+z= 33
If any ofx, y, zare 0, then they all are and therefore, none ofx, y, zare 0 and so alsoλ6= 0. We easily get x=y =z and so 3x= 33, hence x = 11 = y =z. So 113 is the maximum value of xyz.
Then the maximum value ofQni=1xi givenPni=1xi=soccurs when x1 =x2 =· · ·=xn=
Pn i=1xi
n =
s
n. This shows
n Y
i=1
xi ≤
Pn i=1xi
n n
which can be re-written as
n √
x1x2· · ·xn≤
x1+x2+· · ·+xn
n
Exercise 9. Find the absolute extrema off(x, y) =x2+y2+ 2y−2 on (x−1)2+y2≤2. Solution 9. First find critical points (a, b) satisfying (a−1)2+b2<2, then either use LM to find critical points on the boundary - or - parametrize the boundary as a single variable function and use single variable calculus.
Step 1. (Critical points.) Solve
which happens at (0,−1). (This actually occurs on the boundary! Not intended - oops.) Step 2 (a). Use LM. So solve
2x= 2λ(x−1) 2y+ 2 = 2λy (x−1)2+y2 = 2 These give (assumingλ6= 0)
xy= (x−1)(x+ 1)
which reduces tox= 1 +y. So we have 2y2 = 2 andy=±1. This gives (0,−1) and (2,1).
Ifλ= 0, then x= 0 and y+ 1 = 0 so again we have (0,−1).
Step 2 (b). Parametrize: r(t) = (1 +√2 cos(t),√2 sin(t)) so thatf(r(t)) = 2√2(cos(t) + sin(t)) + 1 and f0(t) = 2√2(cos(t)−sin(t)) = 0 when cos(t) = sin(t), at t = π/4 and t= 5π/4. This is (0,−1) and (2,1).
