Solution

f : D \to R

D

is convex when

f ((1 - t) x + t y) \leq (1 - t) f (x) + t f (y)

for all

x, y \in D

and

0 \leq t \leq 1

f

is once differentiable, this is equivalent to monotonicity of its gradient:

(\nabla f (x) - \nabla f (y)) \cdot (x - y) \geq 0 .

Equivalently, every tangent hyperplane supports the graph:

f (y) \geq f (x) + \nabla f (x) \cdot (y - x) .

f

is twice differentiable, convexity is equivalent to the hessian matrix being positive semidefinite throughout

D

For

f (x, y) = x^{3} + y^{3} + A x y,

the Hessian is

H = (6 x A A 6 y) .

A real symmetric two-by-two matrix is positive semidefinite exactly when its two diagonal entries and its determinant are nonnegative. Thus

x \geq 0, y \geq 0, 36 x y - A^{2} \geq 0.

The largest convexity domain is therefore

D_{A} = {(x, y) : x \geq 0, y \geq 0, 36 x y \geq A^{2}} .

For

A \neq = 0

, its boundary is the hyperbola

y = A^{2} / (36 x)

in the first quadrant and the domain lies above it. For

A = 0

Solved by gpt-5.6-sol high.

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