Maximum/Minimum

Question

Find the maximum and minimum values of the function f(x,y)=e^xy subject to x^3+y^3=16

Answer 1

We have a multivariable function subject to a constraint, so the best way to go here is the method of Lagrange multipliers.

Step 1. Let's define the function to be optimized f(x,y) = e^(x*y) and the constraint function g(x,y) = x^3 + y^3 - 16. We now introduce a new function

L(x,y,λ) = f(x,y) + λ*g(x,y),

= e^xy + λ(x³+y³-16)

where λ is an auxiliary variable called the Lagrange multiplier.

Step 2. According to the method of Lagrange multipliers, we now consider the partial derivative of L with respect to x, y, and λ, setting all three partial derivatives to zero. This gives

L_x = ye^xy + 3λx² = 0, (Equation 1)

L_y = xe^xy + 3λy² = 0, (Equation 2)

L_λ = x³ + y³ - 16 = 0, (Equation 3)

which gives us three equations for three unknowns (not a bad place to be).

Step 3. The hard part now is solving the system above, which will involve a variety of algebra tricks. What makes these equations so hard is the term e^(x*y), so what I am going to do is combine Equations 1 and 2 in a way that eliminates this term. Notice this can be accomplished by considering the combination x*(Equation 1) - y*(Equation 2). More explicitly, we have

x*(Equation 1): xye^xy + 3λx³ = 0,

y*(Equation 2): xye^xy + 3λy³ = 0,

x*(Equation 1) - y*(Equation 2): 3λ(x³-y³) = 0.

So I obtain a new equation 3λ(x^3-y^3) = 0. Let's call this Equation 4.

From Equation 4, I deduce that either λ = 0 or x = y. This leads us to consider both cases separately.

Case 1. Suppose λ = 0. Then, Equation 1 implies that y = 0, and Equation 2 implies that x = 0. Given x = y = 0, Equation 3 becomes a contradiction since x^3 + y^3 = 0 =/= 16. Thus, there are no solutions in this case.

Case 2. Suppose x = y. Equation 3 implies then that x = y = 2. If x = y = 2, then Equation 1 (or Equation 2) implies λ = -e^4/6.

To summarize, we have the solution (x = 2, y = 2, λ = -e^4/6) from this case.

Step 4. The fourth and final step of the method (and one that is often forgotten about) is to determine whether the solution found in the previous step is a max or min. Suppose this point was a minimum. It stands to reason then that, for all sufficiently large x and y satisfying Equation 3, f(x,y) > e^4. However, this is not true. As x --> -∞, we see from Equation 3 that y --> +∞ so that f(x,y) = e^(x*y) --> 0. Similar behavior occurs as x --> +∞ so that f(x,y) --> 0 as x and y become sufficiently large in Equation 3. Thus, e^4 must not be a minimum of f(x,y) subject to g(x,y) = 0, but a maximum.

Answer 2

Lagrange function F(x,y,λ)= e^xy - λ(x³ + y³ - 16);

F_x = ye^xy - 3x²λ = 0; F_y = xe^xy - 3y²λ = 0; F_λ = - x³ - y³ + 16 = 0;

y/x = x²/y²; y³ = x³; y = x; 2x³ = 16; x = 2; y = 2. Point (2,2). f(2,2) = e⁴.

To show that at (2, 2) we have maximum. Let take some different point of x³ + y³ = 16,

for example x = 2^4/3, y = 0. f(2^4/3, 0) = e⁰ = 1 < e⁴.

So, max f(x, y) = f(2,2) = e⁴ = 54.5982