Solution to problem from 3/16
1. Maximize xy subject to the constraint x3 + y 3 = 1.
Solution: We’ll use the method of Lagrange multipliers to find the critical points. f (x, y) = xy; our
3
3
constraint can be written asg(x, y) = 0, where
g(x,
y) = x +y −1. So we want points such that 5f = λ5g
2
y
3x
for some value of λ. 5f =
, 5g =
. So we have three equations:
x
3y 2
y = λ3x2
(1)
2
(2)
x +y =1
(3)
x = λ3y
3
3
Solving the first equation for λ, we get λ = 3xy 2 - or x = 0, in which case that step was illegal. We’ll deal
with that case later. Assuming x 6= 0, we have λ = 3xy 2 . Plugging this into the second equation, we have
3
x = 3xy 2 3y 2 . Simplifying, x = xy 2 . Multiplying both sides by x2 , we get x3 = y 3 , so x = y. Plugging this into
1
1
1
the third equation, x3 + x3 = 1, so x3 = 21 , and x = y = 21/3
. That gives us a critical point at ( 21/3
, 21/3
).
What about x = 0? If x = 0, then the first equation says y = 0, but the third equation says y = 1. So
1
1
1
1
1
, 21/3
) is our only critical point. f ( 21/3
, 21/3
) = 22/3
. How do we know
there’s no solution in that case. ( 21/3
if that’s actually a maximum? It could be a minimum, or a saddle.
The critical point is the only point at which the function is allowed to “turn around”. If it’s not a
maximum, then that means that the function must get bigger in some direction, and keep getting bigger
forever. But when x is large, x3 + y 3 = 1 means that y must be negative, so xy < 0. When y is large,
x3 + y 3 = 1 means that x must be negative, so xy < 0. So the function gets smaller in every direction; this
is a maximum.
1