Lagrange multipliers example
March 11, 2011
Problem: Maximize f (x, y) = 2x + 3y on the set S = {(x, y) | x2 + 2y 2 = 34}.
Answer: Define g(x, y) = x2 + 2y 2 , and note that S = g −1 (34). Hence by Theorem 25, we can find the
extrema of f on S by setting
∇f (x, y) = λ∇g(x, y).
2
2x
Since ∇f (x, y) =
and ∇g(x, y) =
, we get the following system of three equations:
3
4y
2 = λ2x
3 = λ4y
2
x + y 2 = 34
We include the third equation because we are only searching for the extrema of f on set S.
Note that x 6= 0, for otherwise the first equation would be false. Hence we can divide by 2x in the first
equation to get x1 = λ. Similarly, y 6= 0, for otherwise the second equation would be false. Hence we can
3
divide by 4y in the second equation to get 4y
= λ.
3
4
1
We have x = λ = 4y , so x = 3 y. Plugging this into the third equation, we get
4 2
34 2
y + 2y 2 = 34 =⇒
y = 34 =⇒ y 2 = 9 =⇒ y = 3 or − 3.
3
9
From the equation x = 34 y, we see that if y = 3 then x = 4, and if y = −3 then x = −4. So our two possible
extrema are (x, y) = (4, 3) and (x, y) = (−4, −3).
Since set S is closed and bounded and function f is continuous, Theorem 24 tells us that f attains a global
maximum on S. A global maximum is also a local maximum, so the global maximum must be either at (4, 3)
or (−4, −3). We compare the values f (4, 3) = 17 and f (−4, −3) = −17 to see that the global maximum of
f on S occurs at (4, 3) and has value 17. Boom!
1