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  2. For business economics
  3. Chapter 5: Optimization
  4. Optimization functions of one variable
  5. Second-order condition extremum
  6. Exercise 2

Exercise 2

Determine all the extrema of $y(x)=-x^3+5x^2+8x+5$.
  • $y(-\frac{2}{3})=2\frac{5}{27}$ is a minimum
  • $y(4)=53$ is a maximum
  • $y(-\frac{2}{3})=14$ is a minimum
  • $y(4)=-14$ is a maximum
  • $y(-\frac{2}{3})=2\frac{5}{27}$ is a maximum
  • $y(4)=53$ is a minimum
  • $y(-\frac{2}{3})=14$ is a maximum
  • $y(4)=-14$ is a minimum
Determine all the extrema of $y(x)=-x^3+5x^2+8x+5$.
Antwoord 1 correct
Correct
Antwoord 2 optie
  • $y(-\frac{2}{3})=14$ is a minimum
  • $y(4)=-14$ is a maximum
Antwoord 2 correct
Fout
Antwoord 3 optie
  • $y(-\frac{2}{3})=2\frac{5}{27}$ is a maximum
  • $y(4)=53$ is a minimum
Antwoord 3 correct
Fout
Antwoord 4 optie
  • $y(-\frac{2}{3})=14$ is a maximum
  • $y(4)=-14$ is a minimum
Antwoord 4 correct
Fout
Antwoord 1 optie
  • $y(-\frac{2}{3})=2\frac{5}{27}$ is a minimum
  • $y(4)=53$ is a maximum
Antwoord 1 feedback
Correct: $y'(x)=-3x^2+10x+8$. $y'(x)=0$ gives $x=-\frac{2}{3}$ and $x=4$. $y''(x)=-6x+10$. Since $y''(-\frac{2}{3})=14>0$ it holds that $y(-\frac{2}{3})=2\frac{5}{27}$ is a minimum, and since $y''(4)=-14<0$ it holds that $y(4)=53$ is a maximum.

Go on.
Antwoord 2 feedback
Wrong: The value of the extremum can be found by the use of the original function, not by the use of the second-order derivative.

Try again.
Antwoord 3 feedback
Wrong: $y''(c)>0$ means that a stationary point $c$ is a minimum location.

See Second-order condition extremum.
Antwoord 4 feedback
Wrong: The value of the extremum can be found by the use of the original function, not by the use of the second-order derivative.

Try again.