5.3 Conditions under which a stationary point is a global optimum
One variable
Let f be a differentiable concave function of a single variable. Then by a previous result, for every point x, no point on the graph of f lies above the tangent to f at x. Thus if f'(x) = 0 (i.e. if x is a stationary point of f), the point x is a (global) maximizer of f. Similarly, a differentiable convex function lies nowhere below any of its tangents. Combining these observations with a previous result, we have the following result. Proposition

Let f be a differentiable function of a single variable defined on the interval I, and let x be in the interior of I.
 If f is concave then x is a global maximizer of f in I if and only if x is a stationary point of f (i.e. f'(x) = 0).
 If f is convex then x is a global minimizer of f in I if and only if x is a stationary point of f (i.e. f'(x) = 0).
 Proof

First suppose that f is concave. If f'(x) = 0 then by a previous result we have f(x') − f(x) ≤ 0 for all x' ∈ I, or f(x') ≤ f(x) for all x' ∈ I. Thus x is a (global) maximizer of f in I. If x is a (global) maximizer of f in the interior of I then f'(x) = 0 by a previous result.
Symmetric arguments apply if f is convex.
Now, a twicedifferentiable function is concave if and only if its second derivative is nonpositive (and similarly for a convex function), so we deduce that if f is a twicedifferentiable function defined on the interval I and x is in the interior of I then
 f"(z) ≤ 0 for all z ∈ I ⇒ [x is a global maximizer of f in I if and only if f'(x) = 0]
 f"(z) ≥ 0 for all z ∈ I ⇒ [x is a global minimizer of f in I if and only if f'(x) = 0].
 Example
 Consider the problem max_{x}−x^{2} subject to x ∈ [−1,1]. The function −x^{2} is concave; its unique stationary point is 0. Thus its global maximizer is 0, which is in [−1,1] and is thus the solution of the problem.
 Example

A firm produces output using a single input. The unit price of the input is w and the unit price of output is p. The firm's output from x units of the input is √x. For what value of x is its profit maximized?
The firm's profit is p√x − wx. The derivative of this function is (1/2)px^{−1/2} − w, and the second derivative is −(1/4)px^{−3/2} ≤ 0 for all x, so the function is concave. So the global maximizer of the function is a stationary point. Hence this maximizer solves (1/2)px^{−1/2} − w = 0, so that x = (p/2w)^{2}.
What happens if the production function is x^{2}?
Many variables
A differentiable concave function of many variables always lies below, or on, its tangent plane, and a differentiable convex function always lies above, or on, its tangent plane. Thus as for functions of a single variable, every stationary point of a concave function of many variables is a maximizer, and every stationary point of a convex function of many variables is a minimizer, as the following result claims. Proposition

Let f be a differentiable function of n variables defined on the convex set S, and let x be in the interior of S.
 If f is concave then x is a global maximizer of f in S if and only if it is a stationary point of f (i.e. f'_{i}(x) = 0 for i = 1, ..., n).
 If f is convex then x is a global minimizer of f in S if and only if it is a stationary point of f (i.e. f'_{i}(x) = 0 for i = 1, ..., n).
 Proof

First suppose that f is concave. If f'_{i}(x) = 0 for i = 1, ..., n then by a previous result we have f(x') − f(x) ≤ 0 for all x' ∈ S, or f(x') ≤ f(x) for all x' ∈ S. Thus x is a (global) maximizer of f in S. If x is a (global) maximizer of f in the interior of S then f'_{i}(x) = 0 for i = 1, ..., n by a previous result.
Symmetric arguments apply if f is convex.
 H(z) is negative semidefinite for all z ∈ S ⇒ [x is a global maximizer of f in S if and only if x is a stationary point of f]
 H(z) is positive semidefinite for all z ∈ S ⇒ [x is a global minimizer of f in S if and only if x is a stationary point of f],
Be sure to notice the difference between the form of this result and that of the the result on local optima. To state briefly the results for maximizers together:
Sufficient conditions for local maximizer: if x* is a stationary point of f and the Hessian of f is negative definite at x* then x* is a local maximizer of f
Sufficient conditions for global maximizer: if x* is a stationary point of f and the Hessian of f is negative semidefinite for all values of x then x* is a global maximizer of f.
 Example

Consider the function f(x, y) = x^{2} + xy + 2y^{2} + 3. We have
f'_{x}(x, y) = 2x + y f'_{y}(x, y) = x + 4y We have
f"_{xx}(x, y) = 2, f"_{yy}(x, y) = 4, and f"_{xy}(x, y) = 1,so the Hessian of f is2 1 1 4 Thus the global minimizer of the function is (0, 0); the minimum value of the function is 3.
 Example

Consider the function f(x, y) = x^{4} + 2y^{2}. We have
f'_{x}(x, y) = 4x^{3} f'_{y}(x, y) = 4y, f"_{xx}(x, y) = 12x^{2}, f"_{yy}(x, y) = 4, and f"_{xy}(x, y) = 0,so the Hessian of f is12x^{2} 0 0 4 .
 Example

Consider the function f(x, y) = x^{3} + 2y^{2}. We have
f'_{x}(x, y) = 3x^{2} f'_{y}(x, y) = 4y, f"_{xx}(x, y) = 6x, f"_{yy}(x, y) = 4, and f"_{xy}(x, y) = 0,so the Hessian of f is6x 0 0 4 . At (x, y) = (0, 0) the matrix is positive semidefinite, but not positive definite. Thus we cannot tell from this analysis whether (0, 0) is a local maximizer or local minimizer, or neither.
We conclude that if f has a minimizer, then this minimizer is (0, 0), and if f has a maximizer, then this maximizer is (0, 0).
In fact, f does not have either a minimizer or a maximizer: for any given value of y, f(x, y) is an arbitrarily large negative number when x is an arbitrarily large negative number, and for any given value of x, f(x, y) is an arbitrarily large positive number when y is an arbitrarily large positive number. Further, (0, 0) is neither a local maximizer nor a local minimizer: for all ε > 0 we have f(ε, 0) > 0 and f(−ε, 0) < 0.
 Example

Consider the function f(x_{1}, x_{2}, x_{3}) = x2
1 + 2x2
2 + 3x2
3 + 2x_{1}x_{2} + 2x_{1}x_{3}, found to be strictly convex in a previous example. It has a unique stationary point, (x_{1}, x_{2}, x_{3}) = (0, 0, 0). Hence its unique global minimizer is (0, 0, 0), with a value of 0.
 Example

Consider a firm with the production function f, defined over vectors (x_{1}, ..., x_{n}) of inputs. Assume that f is concave. The firm's profit function is
π(x_{1}, ..., x_{n}) = pf(x_{1}, ..., x_{n}) −where p is the price of the output of the firm and w_{j} is the price of the jth input. The second term in the profit function is linear, hence concave, and so the function π is concave (by a previous result). So the input bundle (x*_{1}, ..., x*_{n}) > 0 maximizes the firm's profit if and only if it is a stationary point of π, or
∑n
j=1w_{j}x_{j}pf'_{j}(x*_{1}, ..., x*_{n}) = w_{j} for j = 1, ..., n(i.e. the value of the marginal product of each input j is equal to its price).