Area of a box is simply the sum of the areas of each

area of a box is simply the sum of the areas of each of the sides so the constraint is given by
2 xy + 2 yz + 2 xz = 64 = ⇒ xy + yz + xz = 32 .
Note that we divide the constraint by 2 to simplify the equation a little. Also, we get the function g ( x, y, z )
from this.
g ( x, y, z ) = xy + yz + xz.
Here are the four equations that we need to solve.
yz = λ ( y + z ) ( f
x
= λg
x
) , (3.4)
xz = λ ( x + z ) ( f
y
= λg
y
) , (3.5)
xy = λ ( x + y ) ( f
z
= λg
z
) , (3.6)
xy + yz + xz = 32 ( g ( x, y, z ) = 32) . (3.7)
Although the equations are nonlinear, there are many ways to solve this system. We will solve it in the
following way. Let us multiply equation (3.4) by x , equation (3.5) by y and equation (3.6) by z .
xyz = λx ( y + z ) , (3.8)
xyz = λy ( x + z ) , (3.9)
xyz = λz ( x + y ) . (3.10)
Now notice that we can set equations (3.8) and (3.9) equal. Doing this gives
λx ( y + z ) = λy ( x + z ) ,
λ ( xy + xz ) - λ ( xy + yz ) = 0 ,
λ ( xz - yz ) = 0 = ⇒ λ = 0 or xz = yz.
This implies two possibilities. The first, λ = 0, is not possible since if this is the case equation (3.4) will
reduce to
yz = 0 = ⇒ y = 0 or z = 0 .
Since we are talking about the dimensions of a box neither of these are possible so we can discount λ = 0.
This leaves the second possibility.
xz = yz.
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3. Partial Derivatives
Since we know that z 6 = 0 (again since we are talking about the dimensions of a box) we can cancel the z
from both sides. This gives
x = y. (3.11)
Next, let us set equations (3.9) and (3.10) equal. Doing this gives
λy ( x + z ) = λz ( x + y ) ,
λ ( xy + yz - xz - yz ) = 0 ,
λ ( xy - xz ) = 0 = ⇒ λ = 0 or xy = xz.
As already discussed we know that λ = 0 won’t work and so this gives
xy = xz.
We can also say that x 6 = 0 since we are dealing with the dimensions of a box so we have
y = z. (3.12)
Plugging equations (3.11) and (3.12) into equation (3.7) we get
y
2
+ y
2
+ y
2
= 3 y
2
= 32 , y = ±
r
32
3
≈ ± 3 . 266 .
However, we know that y must be positive since we are talking about the dimensions of a box. Therefore
the only solution that makes physical sense here is
x = y = z ≈ 3 . 266 cm .
This shows that we have a cube here.
We should be a little careful here. Since we have obtained only one solution we might be tempted to
assume that this is the dimensions that will give the largest volume. The method of Lagrange multipliers
will give a set of points that will either maximize or minimize a given function subject to the constraint.
However, when we get a single solution it may be either a maximum or a minimum. To verify that we indeed
have a maximum, as we want, all we need to do is pick any other point that satisfies the constraint and
check its volume against the volume of the point we got above. If the volume of the point above is larger
than the second point we will know that we indeed have a maximum.
To get the second point let us choose y = z = 2 plugging these into the constraint gives
2 x + 2 x + 4 = 32 , x = 7 .
Checking the volume at the two points gives
f (3 . 266 , 3 . 266 , 3 . 266) = 34 . 8376 ,
f (7 , 2 , 2) = 28 .