Generating...                               quiz05c_n19

  1. Let $f(x,y) = x^2+x^3+2\,y^2+x\,y^2$. Find the value $D$ at the critical point $\displaystyle \left[ y=0 , x=-\frac{2}{3} \right] $ for second derivative test.

    $D = -\frac{16}{3}$ $D = 24$ $D = -\frac{40}{3}$ $D = 8$

  2. Find the extreme value for $f(x,y,z) = 3\,x^2+y^2+z^2$ subject to x + y + 3z = 2 .

    The extreme value $\displaystyle \frac{27}{31}$ at $\displaystyle \left[ z=\frac{27}{31} , y=\frac{9}{31} , x=\frac{3}{31} \right] $

    The extreme value $\displaystyle \frac{9}{11}$ at $\displaystyle \left[ z=\frac{9}{11} , y=\frac{3}{11} , x=\frac{3}{11} \right] $

    The extreme value $\displaystyle \frac{12}{31}$ at $\displaystyle \left[ z=\frac{18}{31} , y=\frac{6}{31} , x=\frac{2}{31} \right] $

    The extreme value $\displaystyle \frac{4}{11}$ at $\displaystyle \left[ z=\frac{6}{11} , y=\frac{2}{11} , x=\frac{2}{11} \right] $

  3. Choose the correct answer regarding the critical point [ y = 0 ,  x = 0 ] for $f(x,y) = x^2+x^3+3\,y^2+x\,y^2$.

    Since $D = 12$ and $f_{xx} = 2$, it is a local minimum

    Since $D = 12$ and $f_{xx} = 2$, it is a saddle point

    Since $D = -\frac{28}{3}$ and $f_{xx} = -2$, it is a saddle point

    Since $D = -\frac{28}{3}$ and $f_{xx} = -2$, it is a local maximum

  4. Suppose that $\displaystyle -\frac{4\,x}{49}-\frac{9\,y}{49}+\frac{36\,z}{49}=1$ is the tangent plane at the point [  − 1 ,   − 1 ,  1 ] to the ellipsoid $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1$ . Then find the values $a$, $b$ and $c$.

    $a = \frac{7}{2}$, $b = \frac{7}{3}$, and $c = \frac{7}{6}$. $a = \frac{7}{3}$, $b = \frac{7}{2}$, and $c = \frac{7}{6}$. $a = \frac{7}{6}$, $b = \frac{7}{3}$, and $c = \frac{7}{2}$. $a = \frac{7}{2}$, $b = \frac{7}{6}$, and $c = \frac{7}{3}$.

  5. Let $f(x,y) = 2\,x^2+x^3+2\,y^2+x\,y^2$. Find the gradient $\nabla f(x,y)$ .

    $\displaystyle \left[ 2\,x+3\,x^2+y^2 , 6\,y+2\,x\,y \right] $

    $\displaystyle \left[ 2\,x+3\,x^2+y^2 , 4\,y+2\,x\,y \right] $

    $\displaystyle \left[ 4\,x+3\,x^2+y^2 , 4\,y+2\,x\,y \right] $

    $\displaystyle \left[ 4\,x+3\,x^2+y^2 , 6\,y+2\,x\,y \right] $

  6. Choose the correct statement regarding the critical points for $f(x,y) = 2\,x^2+x^3+3\,y^2+x\,y^2$.

    The critical points $(x,y)$ satisfy \begin{displaymath}\begin{cases} 2\,x+3\,x^2+y^2 = 0; \\ 4\,y+2\,x\,y = 0. \end{cases}\end{displaymath}

    The critical points $(x,y)$ satisfy \begin{displaymath}\begin{cases} 4\,x+3\,x^2+y^2 = 0; \\ 4\,y+2\,x\,y = 0. \end{cases}\end{displaymath}

    The critical points $(x,y)$ satisfy \begin{displaymath}\begin{cases} 4\,x+3\,x^2+y^2 = 0; \\ 6\,y+2\,x\,y = 0. \end{cases}\end{displaymath}

    The critical points $(x,y)$ satisfy \begin{displaymath}\begin{cases} 2\,x+3\,x^2+y^2 = 0; \\ 6\,y+2\,x\,y = 0. \end{cases}\end{displaymath}

  7. Let $\displaystyle \frac{4\,x^2}{49}+\frac{9\,y^2}{49}+\frac{36\,z^2}{49}=1$ be the ellipsoid. Find the tangent plane at the point [ 1 ,   − 1 ,  1 ] .

    $\displaystyle \frac{4\,x}{49}-\frac{36\,y}{49}+\frac{9\,z}{49}=1$ $\displaystyle \frac{9\,x}{49}-\frac{36\,y}{49}+\frac{4\,z}{49}=1$ $\displaystyle \frac{4\,x}{49}-\frac{9\,y}{49}+\frac{36\,z}{49}=1$ $\displaystyle \frac{9\,x}{49}-\frac{4\,y}{49}+\frac{36\,z}{49}=1$

  8. Choose the correct answer regarding the critical point [ y = 0 ,  x = 0 ] for $f(x,y) = 2\,x^2+x^3+2\,y^2+x\,y^2$.

    It is a local minimum

    It is a saddle point

  9. Find the extreme value for $f(x,y) = e^{2\,x\,y}$ subject to $\displaystyle x^3+y^3=2$ .

    The extreme value $\displaystyle e^2$ at x = 1 and y = 1

    The extreme value $\displaystyle e^{18}$ at x = 3 and y = 3

    The extreme value e at x = 1 and y = 1

    The extreme value $\displaystyle e^9$ at x = 3 and y = 3

  10. Let $f(x,y) = 2\,x^2+2\,x^2\,y+y^2$. Find the value $D$ at the critical point [ y = −1 ,  x =  − 1 ] for second derivative test.

    $D = -8$ $D = 8$ $D = -16$ $D = 4$


Department of Mathematics
Last modified: 2025-09-14