- Maximize $z(x,y)=3x^2y$
- Subject to $(x+1)y=10$
- Where $x,y<-1$
Antwoord 1 correct
Correct
Antwoord 2 optie
$z(0,10)=0$
Antwoord 2 correct
Fout
Antwoord 3 optie
$z(-\frac{115}{56},-\frac{560}{59})=-120$
Antwoord 3 correct
Fout
Antwoord 4 optie
$z(-5,-2\frac{1}{2})=187\frac{1}{2}$
Antwoord 4 correct
Fout
Antwoord 1 optie
$z(-2,-10)=-120$
Antwoord 1 feedback
Correct: $y=\dfrac{10}{x+1}$.
$Z(x)=\dfrac{30x^2}{x+1}$.
$Z'(x)=\dfrac{30x(x+2)}{(x+1)^2}$. Hence, $Z'(x)=0$ for $x=0$ (outside domain) or $x=-2$. Via a sign chart you can determine that $x=-2$ is a maximum location.
$y(-2)=\dfrac{10}{-2+1}=-10$.
Hence, $z(-2,-10)=-120$ is the solution.
Go on.
$Z(x)=\dfrac{30x^2}{x+1}$.
$Z'(x)=\dfrac{30x(x+2)}{(x+1)^2}$. Hence, $Z'(x)=0$ for $x=0$ (outside domain) or $x=-2$. Via a sign chart you can determine that $x=-2$ is a maximum location.
$y(-2)=\dfrac{10}{-2+1}=-10$.
Hence, $z(-2,-10)=-120$ is the solution.
Go on.
Antwoord 2 feedback
Wrong: $(x,y)=(0,10)$ is not part of the domain.
Try again.
Try again.
Antwoord 3 feedback
Wrong: You cannot find the zeros of $\dfrac{30x^2+60x}{(x+1)^2}$ by putting the nominator and denominator equal to each other. Moreover, you cannot round intermediate solutions ($\sqrt{3252}$).
Try again.
Try again.
Antwoord 4 feedback
Wrong: $z(-5,-2\frac{1}{2})\neq187\frac{1}{2}$, because $(-5)^2\neq -25$.
Try again.
Try again.