MA22S1: SOLUTIONS TO TUTORIAL 5
1. Find all critical points of the function f (x, y) = x3 + 3xy + y 3 and
classify each critical point as a local maximum, local minimum or
saddle point.
Solution: First note that since f (x, y) is a polynomial in two variables, all partial derivatives of f (x, y) exist and are continuous at every point in R2 . Also since the domain of f (x, y) is all of R2 there are
no boundary points to check. Thus by the 1st Derivatives Test, critical points for f (x, y) can only occur where fx (x, y) = fy (x, y) = 0.
We have
fx (x, y) = 3x2 + 3y = 0 =⇒ y = −x2
fy (x, y) = 3x + 3y 2 = 0 =⇒ x = −y 2
Thus
x = −(−x2 )2 = −x4 =⇒ x(x3 + 1) = 0
and so x = 0 or x = −1.
If x = 0 then y = 0 and so (0, 0) is a critical point for f (x, y).
If x = −1 then y = −1 and so (−1, −1) is a critical point for f (x, y).
To classify these two critical points we use the 2nd Derivatives
Test. We have
fxx (x, y) = 6x,
fxy (x, y) = 3,
1
fyy (x, y) = 6y
2
MA22S1: SOLUTIONS TO TUTORIAL 5
x*x*x+3*x*y+y*y*y
140
120
100
80
60
40
20
0
-20
3
-40
2
1
-60
-3
-2
0
-1
x
0
-1
1
y
-2
2
3
-3
Figure 1. Graph of f (x, y) = x3 + 3xy + y 3
Thus at (0, 0),
fxx (0, 0)fyy (0, 0) − fxy (0, 0)2 = −9 < 0 =⇒ saddle point.
At (−1, −1) we have,
fxx (−1, −1) = −6 < 0
fxx (−1, −1)fyy (−1, −1)−fxy (−1, −1)2 = −6(−6)−9 > 0 =⇒ local max.
See Figure 1.
2. Use the method of Lagrange multipliers to find the maximum volume
of a rectangular box with surface area 18m2 .
MA22S1: SOLUTIONS TO TUTORIAL 5
3
z
y
x
Figure 2. Rectangular box with sides of length x, y and z
Solution: Suppose the sides of the box have length x, y, z as shown
in Figure 2.
The volume of the box is V (x, y, z) = xyz. The total surface area
is obtained by adding the areas of each side of the box. This means
the surface area is
g(x, y, z) = 2xy + 2yz + 2xz.
We need to find a maximum value for the volume function
V (x, y, z) = xyz
subject to the constraint
g(x, y, z) = 18.
By the method of Lagrange multipliers we seek values x, y, z and
λ which solve the following equations:
∇V (x, y, z) = λ∇g(x, y, z)
and
g(x, y, z) = 18.
We have
∇V (x, y, z) = hVx , Vy , Vz i = hyz, xz, xyi
4
MA22S1: SOLUTIONS TO TUTORIAL 5
∇g(x, y, z) = hgx , gy , gz i = h2(y + z), 2(x + z), 2(x + y)i.
Setting ∇V (x, y, z) = λ∇g(x, y, z) we obtain three equations:
yz = 2λ(y + z),
xz = 2λ(x + z),
xy = 2λ(x + y)
From these equations we obtain
xyz = 2λx(y + z),
xyz = 2λy(x + z),
xyz = 2λz(x + y)
Thus we have
2λx(y + z) = 2λy(x + z) = 2λz(x + y)
and so either λ = 0 or
x(y + z) = y(x + z) = z(x + y).
If λ = 0 then yx = xz = xy = 0 and so g(x, y, z) = 0. But this
contradicts our assumption that g(x, y, z) = 18 so we must have
λ 6= 0. Solving the equations x(y + z) = y(x + z) = z(x + y) we have
x(y + z) = y(x + z) =⇒ xz = yz =⇒ x = y
x(y + z) = z(x + y) =⇒ xy = yz =⇒ x = z
and so x = y = z. (Note that we assume here that each side of the
box has non-zero length so x, y, z 6= 0). Thus
g(x, y, z) = 6x2 = 18
and so x = y = z =
√
3. The maximum volume of the box is
√ √ √
√
V ( 3, 3, 3) = 3 3 m3 .