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  2. For business economics
  3. Chapter 5: Optimization
  4. Optimization constrained extremum problems
  5. Method of substitution
  6. Exercise 3

Exercise 3

  • Maximize $z(x,y)=xy-x^3$                      
  • Subject to $\sqrt{y}+x=8$
  • Where $x,y\geq 0$ 
$z(2,36)=64$
$z(1\frac{1}{3},6\frac{2}{3})=6\frac{14}{27}$
$z(0,64)=0$
$z(16,64)=-3072$
  • Maximize $z(x,y)=xy-x^3$                      
  • Subject to $\sqrt{y}+x=8$
  • Where $x,y\geq 0$ 
Antwoord 1 correct
Correct
Antwoord 2 optie
$z(1\frac{1}{3},6\frac{2}{3})=6\frac{14}{27}$
Antwoord 2 correct
Fout
Antwoord 3 optie
$z(0,64)=0$
Antwoord 3 correct
Fout
Antwoord 4 optie
$z(16,64)=-3072$
Antwoord 4 correct
Fout
Antwoord 1 optie
$z(2,36)=64$
Antwoord 1 feedback
Correct: We rewrite $\sqrt{y}+x=8$ to $y=(8-x)^2$. We plug this into the objective function: $Z(x)=x(8-x)^2-x^3=-16x^2+64x$. $Z'(x)=-32x+64$. Putting equal to zero results in $x=2$ and $y=36$. $z(2,36)=64$. We check the boundaries: $z(8,0)=-512$ and $z(0,64)=0$. Hence, $z(2,36)=64$ is a maximum.

Go on.
Antwoord 2 feedback
Wrong: $\sqrt{y}+x=8$ cannot be rewritten as $y=8-x$.

Try again.
Antwoord 3 feedback
Wrong: This is not a maximum, but a minimum.

Try again.
Antwoord 4 feedback
Wrong: To find a stationary point of the function you need to put the derivative (and not the function itself) equal to zero.

See Stationary point.