If a function of one variable is continuous on an interval and has only one critical number, then a local maximum has to be an absolute maximum. But this is not true for functions of two variables. Show that the function
f(x,y)= 3xe^y − x^3 − e^(3y
)has exactly one critical point, and that f has a local maximum there that is not an absolute maximum.
To show that the function f(x, y) = 3xe^y − x^3 − e^(3y) has exactly one critical point, we need to find where the partial derivatives with respect to x and y equal to zero.
1. Find the partial derivative with respect to x:
∂f/∂x = 3e^y - 3x^2
2. Set the partial derivative equal to zero and solve for x:
3e^y - 3x^2 = 0
3e^y = 3x^2
x^2 = e^y
x = ±√(e^y)
3. Find the partial derivative with respect to y:
∂f/∂y = 3xe^y - 3e^(3y)
4. Set the partial derivative equal to zero and solve for y:
3xe^y - 3e^(3y) = 0
e^y(3x - 3e^(2y)) = 0
This equation can have two cases:
a. e^y = 0, which is not possible since e^y is always positive.
b. 3x - 3e^(2y) = 0, which simplifies to x = e^(2y)
Now, we have two critical points on the function:
1. At (x = √(e^y), y)
2. At (x = -√(e^y), y)
However, we need to check if these critical points are local maximums. To do that, we'll find the second partial derivatives and analyze the determinant (D) and the Hessian matrix.
1. Find the second partial derivatives:
∂²f/∂x² = -6x
∂²f/∂y² = 3x - 9e^(3y)
∂²f/∂x∂y = 3e^y
2. Construct the Hessian matrix:
H = [∂²f/∂x², ∂²f/∂x∂y]
[∂²f/∂x∂y, ∂²f/∂y²]
Substituting the second partial derivatives into the Hessian matrix, we get:
H = [-6x, 3e^y]
[3e^y, 3x - 9e^(3y)]
3. Calculate the determinant, D = ∂²f/∂x² * ∂²f/∂y² - (∂²f/∂x∂y)²:
D = (-6x)(3x - 9e^(3y)) - (3e^y)(3e^y)
D = -18x^2 + 54xe^(3y) - 9e^(2y)
Now let's analyze the cases:
1. At (x = √(e^y), y):
Evaluate D at this point:
D = -18(√(e^y))^2 + 54√(e^y)e^(3y) - 9e^(2y)
D = -18e^y + 54√(e^y)e^(3y) - 9e^(2y)
Since D is negative, we have a saddle point at this critical point.
2. At (x = -√(e^y), y):
Evaluate D at this point:
D = -18(-√(e^y))^2 + 54(-√(e^y))e^(3y) - 9e^(2y)
D = -18e^y + 54(-√(e^y))e^(3y) - 9e^(2y)
Since D is positive, we have a local maximum at this critical point.
Therefore, the function f(x, y) = 3xe^y − x^3 − e^(3y) has exactly one critical point at (x = -√(e^y), y), and it has a local maximum at this point that is not an absolute maximum.