# Exercise 3

For which values of $a$ is the function $z(x,y)=ax^2+xy+y^2$ convex?
$a\geq \frac{1}{4}$
For no value of $a$.
$a\geq 0$
$-1\leq a \leq \frac{1}{2}$
For which values of $a$ is the function $z(x,y)=ax^2+xy+y^2$ convex?
Antwoord 1 correct
Correct
Antwoord 2 optie
For no value of $a$.
Antwoord 2 correct
Fout
Antwoord 3 optie
$a\geq 0$
Antwoord 3 correct
Fout
Antwoord 4 optie
$-1\leq a \leq \frac{1}{2}$
Antwoord 4 correct
Fout
Antwoord 1 optie
$a\geq \frac{1}{4}$
Antwoord 1 feedback
Correct: The criterion function $C(x,y)=4a-1$ and the second-order partial derivatives $z''_{xx}(x,y)=2a$ and $z''_{yy}(x,y)=2$ are non-negative for all $a\geq \frac{1}{4}$.

Go on.
Antwoord 2 feedback
Wrong: Reconsider the criterion function of $z(x,y)$.

See second-order condition.
Antwoord 3 feedback
Wrong: For some values of $a\geq 0$ the criterion function $C(x,y)=4a-1$ is negative.

See second-order condition.
Antwoord 4 feedback
Wrong: Reconsider the criterion function of $y(x)$ and the second-order partial derivatives.

See second-order condition.