Generating...                               quiz05c_n1

  1. Choose the correct statement regarding the extreme value for $f(x,y) = e^{3\,x\,y}$ subject to $\displaystyle x^3+y^3=2$ .

    The point $(x,y)$ at the extreme value satisfies \begin{displaymath}\begin{cases} 3\,y\,e^{3\,x\,y} = \lambda ( 3\,x^2); \\ 3\... ...^{3\,x\,y} = \lambda ( 3\,y^2); \\ x^3+y^3 = 2. \end{cases}\end{displaymath}

    The point $(x,y)$ at the extreme value satisfies \begin{displaymath}\begin{cases} e^{3\,x\,y} = \lambda ( 3\,x^2); \\ e^{3\,x\,y} = \lambda ( 3\,y^2); \\ x^3+y^3 = 2. \end{cases}\end{displaymath}

    The point $(x,y)$ at the extreme value satisfies \begin{displaymath}\begin{cases} 3\,y\,e^{3\,x\,y} = \lambda ( x^3+y^3); \\ 3... ...{3\,x\,y} = \lambda ( x^3+y^3); \\ x^3+y^3 = 2. \end{cases}\end{displaymath}

    The point $(x,y)$ at the extreme value satisfies \begin{displaymath}\begin{cases} 3\,x\,y\,e^{3\,x\,y} = \lambda ( 3\,y^2); \\ ... ...^{3\,x\,y} = \lambda ( 3\,x^2); \\ x^3+y^3 = 2. \end{cases}\end{displaymath}

  2. Suppose that $\displaystyle \frac{9\,x}{92}+\frac{9\,y}{23}-\frac{z}{46}=1$ is the tangent plane at the point [ 2 ,  2 ,   − 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 $c$.

    $a = \frac{\sqrt{46}}{3}$, $b = \frac{2\,\sqrt{46}}{3}$, and $c = \sqrt{46}$. $a = \frac{\sqrt{46}}{3}$, $b = \sqrt{46}$, and $c = \frac{2\,\sqrt{46}}{3}$. $a = \sqrt{46}$, $b = \frac{\sqrt{46}}{3}$, and $c = \frac{2\,\sqrt{46}}{3}$. $a = \frac{2\,\sqrt{46}}{3}$, $b = \frac{\sqrt{46}}{3}$, and $c = \sqrt{46}$.

  3. 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} 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; \\ 4\,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}

    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}

  4. Let $f(x,y) = 2\,x^2+x^3+2\,y^2+x\,y^2$. Find the value $D$ at the critical point $\displaystyle \left[ y=0 , x=-\frac{4}{3} \right] $ for second derivative test.

    $D = 16$ $D = -\frac{28}{3}$ $D = -\frac{16}{3}$ $D = 12$

  5. Let $\displaystyle \frac{4\,x^2}{157}+\frac{9\,y^2}{157}+\frac{36\,z^2}{157}=1$ be the ellipsoid. Find the tangent plane at the point [ 1 ,  1 ,  2 ] .

    $\displaystyle \frac{4\,x}{157}+\frac{36\,y}{157}+\frac{18\,z}{157}=1$ $\displaystyle \frac{9\,x}{157}+\frac{36\,y}{157}+\frac{8\,z}{157}=1$ $\displaystyle \frac{4\,x}{157}+\frac{9\,y}{157}+\frac{72\,z}{157}=1$ $\displaystyle \frac{36\,x}{157}+\frac{9\,y}{157}+\frac{8\,z}{157}=1$

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

    It is a saddle point It is a local minimum

  7. Let $f(x,y) = x^2+x^2\,y+y^2$. Find the value $D$ at the critical point [ y = 0 ,  x = 0 ] for second derivative test.

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

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

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

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

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

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

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

    Since $D = 16$ and $f_{xx} = 4$, it is a local minimum

    Since $D = 16$ and $f_{xx} = 4$, it is a saddle point

    Since $D = -\frac{16}{3}$ and $f_{xx} = -4$, it is a local maximum

    Since $D = -\frac{16}{3}$ and $f_{xx} = -4$, it is a saddle point

  10. Let $f(x,y) = x^2+2\,x^2\,y+y^2$. Find all the critical points.

    [ y = 0 ,  x = 0 ] , $\displaystyle \left[ y=-\frac{1}{2} , x=-\frac{1}{\sqrt{2}} \right] $ and $\displaystyle \left[ y=-\frac{1}{2} , x=\frac{1}{\sqrt{2}} \right] $ [ y = 0 ,  x = 0 ] , [ y = −1 ,  x =  − 1 ] and [ y = −1 ,  x = 1 ] [ y = 0 ,  x = 0 ] , $\displaystyle \left[ y=-1 , x=-\sqrt{2} \right] $ and $\displaystyle \left[ y=-1 , x=\sqrt{2} \right] $ [ y = 0 ,  x = 0 ] , [ y = −2 ,  x = 2 ] and [ y = −2 ,  x =  − 2 ]


Department of Mathematics
Last modified: 2025-06-19