1. Home
  2. For business economics
  3. Chapter 5: Optimization
  4. Optimization convex/concave functions
  5. Functions of two variables
  6. Extrema and saddle points
  7. Exercise 1

Exercise 1

Determine all the extremum locations of the function $z(x,y)=x^2+xy+y^2$.
The point $(0,0)$ is a minimum location.
The function has no extrema.
The point $(0,0)$ is a maximum location.
The point $(-1,-1)$ is a minimum location and the point $(1,1)$ is a maximum location.
Determine all the extremum locations of the function $z(x,y)=x^2+xy+y^2$.
Antwoord 1 correct
Correct
Antwoord 2 optie
The function has no extrema.
Antwoord 2 correct
Fout
Antwoord 3 optie
The point $(0,0)$ is a maximum location.
Antwoord 3 correct
Fout
Antwoord 4 optie
The point $(-1,-1)$ is a minimum location and the point $(1,1)$ is a maximum location.
Antwoord 4 correct
Fout
Antwoord 1 optie
The point $(0,0)$ is a minimum location.
Antwoord 1 feedback
Correct: The first-order partial derivatives are $z'_x(x,y)=2x+y$ and $z'_y(x,y)=x+2y$. The only point $(x,y)$ that satisfies $2x+y=0$ and $x+2y=0$ is $(x,y)=(0,0)$. From $C(0,0)=3>0$, $z''_{xx}(0,0)=z''_{yy}(0,0)=2>0$ it follows that the point $(0,0)$ is a minimum location.

Go on.
Antwoord 2 feedback
Wrong: Consider the points $(x,y)$ such that $z'_x(x,y)=0$ and $z'_y(x,y)=0$.

Try again.
Antwoord 3 feedback
Wrong: Reconsider the values of the criterion function and the second-order partial derivatives.

See Second-order condition.
Antwoord 4 feedback
Wrong: Do the points $(-1,-1)$ and $(1,1)$ satisfy $z'_x(x,y)=0$ and $z'_y(x,y)=0$?

Try again.