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Math Help - Max and Min

  1. #1
    Junior Member
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    Sep 2008
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    Max and Min

    z=f(x,y)=(y-x^2)(y-2x^2)
    a) Show that the point (0,0) is a critical point of f ' and that the second derivative test for a function of two variables fails at this critical point
    b) Show that f has neither a maximum nor a minimum at (0,0) by proving that there are values of the parameter a for which f has a maximum at (0,0) along the parabolas y=ax^2, while for other values of parameter a, the function f has a minimum at (0,0) along the parabolas y=ax^2.
    c) Show that f has a minimum at (0,0) along every straight line through(0,0)

    This is really urgent to me, i need this for my test
    Can anyone show me how to do? Thanks very much!
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  2. #2
    Junior Member
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    Nov 2008
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    Well I hope it's not too late for your test but here goes

    a) First thing I'd do is develop f so that the derivatives become easier, thus

    <br />
f(x,y)=y^2-3x^2y+2x^4<br />

    therefore

    <br />
f'(x,y)=\begin{matrix} (-6xy+8x^3 & 2y-3x^2) \end{matrix}<br />

    <br />
f''(x,y)=\begin{pmatrix} -6y+24x^2 & -6x \\ -6x & 2 \end{pmatrix}<br />

    So from this we see that f'(x,y)=0, so f is a critical point. However |f''(0,0)|=0, so the second derivative test fails.

    b) let y=ax^2, thus f(x,y) becomes:

    <br />
f(x)=(ax^2-x^2)(ax^2-2x^2)<br />

    <br />
f(x)=(a-1)(a-2)x^4<br />

    So, if a>2 or a<1, f obviously has a min. at x=0 (y is also 0 when x is 0). However if 1<a<2 then f has a max at x=0. Thus, f(x,y) has neither a max or min at (0,0)

    b) let y=mx, where m is any real number. This is the family of straight lines through (0,0). f then becomes:

    <br />
f(x)=m^2x^2-3mx^3+2x^4<br />

    <br />
f'(x)=2m^2x-9mx^2+8x^3<br />

    <br />
f''(x)=2m^2-18mx+24x^2<br />

    So, f'(0)=0, so it's a critical point, as we'd expect. Also, we have that

    <br />
f''(0)=2m^2<br />

    This is positive, and thus f has a min, for all values of m except for zero. If m=0, then,

    <br />
f(x)=2x^4<br />

    In this case f obviously has a min at x=0. The only other case to examine is the straight line x=0, in that case, f is simply f(y)=y^2, which also has an obvious min at (0,0). Thus, f has a min at (0,0) along any straight line passing through (0,0).

    Hope that helps.
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  3. #3
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    thanks so much, you save my life
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