Chapter 4 – Quiz 1 – E600 Mathematics

1. When can you be sure to have found the unique solution of an unconstrained maximization problem with objective $f$ ? (Multiple choice)

When the Hessian of $f$ is indefinite everywhere.

When $f$ is twice continuously differentiable, concave and $f$ has exactly one critical point.

When all but one (interior) critical points of $f$ can be identified as strict local minimizers.

When there is exactly one boundary point $x^b$ , and (i) $f(x^b) = 0$ and (ii) $f(x) < 0$ for all interior points $x\in int(dom(f))$ .

When just one of the (interior) critical points meets the necessary conditions of a local maximizer.

2. Select the requirements of the Weierstrass Extreme Value theorem, which together are a sufficient condition for existence of global extremizers in the unconstrained problem. (Multiple choice)

Non-negativity of the objective function.

Continuity of the objective function.

Closedness of the objective function’s domain.

Convexity of the objective function’s domain.

Boundedness of the objective function’s domain.

Twice-differentiability of the objective function.

Compactness of the objective function’s codomain.

3. What is the global maximizer of $f:\mathbb R\mapsto\mathbb R, x\mapsto x-x^2$ ?

0.5

0.25

There isn’t one.

4. Suppose you are concerned with finding the global maximizer(s) in an unconstrained optimization problem with objective function $f:X\mapsto\mathbb R$ that is twice differentiable. Which candidates $x^*$ can not be ruled out as a potential solution by the first and second order conditions? (Multiple choice)

Boundary points.

Points at which the gradient of $f$ is unequal to the zero vector.

Points at which the Hessian of $f$ is positive definite.

Points at which the Hessian of $f$ is negative semi-definite.

5. Which of the following statements are true for any function $f:X\mapsto\mathbb R$ when $X\subseteq\mathbb R$ ? (Multiple choice)

Any local maximizer of $f$ is also a global maximizer.

If $f$ has a strict global maximizer, then $\arg\max_{x\in X}f(x)$ contains exactly one element.

There might not be a global minimum of $f$ even when $f$ is continuous and $X$ is closed and bounded.

Any global maximizer of $f$ is also a local maximizer.

If existent, $\max X$ is real number.

Whenever $X$ is bounded, $f$ is guaranteed to have a global maximum.

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