Generating...                               quiz05c_n6

  1. Choose the correct answer regarding the critical point $\displaystyle \left[ y=0 , x=-\frac{2}{3} \right] $ for $f(x,y) = x^2+x^3+2\,y^2+x\,y^2$.

    It is a saddle point

    It is a local maximum

  2. Suppose that $\displaystyle \frac{8\,x}{61}-\frac{9\,y}{61}+\frac{36\,z}{61}=1$ is the tangent plane at the point [ 2 ,   − 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 $c$.

    $a = \frac{\sqrt{61}}{3}$, $b = \frac{\sqrt{61}}{2}$, and $c = \frac{\sqrt{61}}{6}$. $a = \frac{\sqrt{61}}{2}$, $b = \frac{\sqrt{61}}{3}$, and $c = \frac{\sqrt{61}}{6}$. $a = \frac{\sqrt{61}}{2}$, $b = \frac{\sqrt{61}}{6}$, and $c = \frac{\sqrt{61}}{3}$. $a = \frac{\sqrt{61}}{3}$, $b = \frac{\sqrt{61}}{6}$, and $c = \frac{\sqrt{61}}{2}$.

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

    [ 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 ] , $\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 = −2 ,  x = 2 ] and [ y = −2 ,  x =  − 2 ]

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

    The extreme value $\displaystyle \frac{9}{4}$ at $\displaystyle \left[ y=-\frac{1}{4} , z=\frac{1}{2} , x=1 \right] $

    The extreme value $\displaystyle \frac{3}{2}$ at $\displaystyle \left[ y=\frac{1}{2} , z=\frac{1}{2} , x=\frac{1}{2} \right] $

    The extreme value $\displaystyle \frac{7}{2}$ at $\displaystyle \left[ y=\frac{5}{2} , z=\frac{1}{2} , x=\frac{1}{2} \right] $

    The extreme value $\displaystyle \frac{17}{4}$ at $\displaystyle \left[ y=\frac{7}{4} , z=\frac{1}{2} , x=1 \right] $

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

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

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

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

  7. Find the extreme value for $f(x,y) = e^{2\,x\,y}$ subject to $\displaystyle x^3+y^3=16$ .

    The extreme value $\displaystyle e^{12}$ at x = 2 and y = 2

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

    The extreme value $\displaystyle e^8$ at x = 2 and y = 2

    The extreme value $\displaystyle e^3$ at x = 1 and y = 1

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

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

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

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

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

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

    The extreme value $\displaystyle \frac{2}{5}$ at $\displaystyle \left[ z=\frac{2}{5} , y=\frac{2}{5} , x=\frac{1}{5} \right] $

    The extreme value $\displaystyle \frac{18}{5}$ at $\displaystyle \left[ z=\frac{6}{5} , y=\frac{6}{5} , x=\frac{3}{5} \right] $

    The extreme value $\displaystyle \frac{1}{3}$ at $\displaystyle \left[ z=\frac{1}{3} , y=\frac{1}{3} , x=\frac{1}{3} \right] $

    The extreme value 3 at [ z = 1 ,  y = 1 ,  x = 1 ]

  10. Let $\displaystyle \frac{36\,x^2}{61}+\frac{4\,y^2}{61}+\frac{9\,z^2}{61}=1$ be the ellipsoid. Find the tangent plane at the point [ 1 ,  2 ,  1 ] .

    $\displaystyle \frac{9\,x}{61}+\frac{72\,y}{61}+\frac{4\,z}{61}=1$ $\displaystyle \frac{9\,x}{61}+\frac{8\,y}{61}+\frac{36\,z}{61}=1$ $\displaystyle \frac{36\,x}{61}+\frac{8\,y}{61}+\frac{9\,z}{61}=1$ $\displaystyle \frac{4\,x}{61}+\frac{72\,y}{61}+\frac{9\,z}{61}=1$


Department of Mathematics
Last modified: 2026-07-16