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Math Help - Maximum and Minimum Problems

  1. #1
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    Maximum and Minimum Problems

    The question statement indicates that the area will be a maximum, but in my attempt you can see that I have arrived at a positive value of 6, suggesting a minimum. I think this has something to do with how I've applied the modulus when calculating the area of the triangle. Specifically, I 'probably' can't define \frac{1}{2}|x^3-6x^2| as either \frac{x^3-6x^2}{2} or \frac{-x^3+6x^2}{2}. Furthermore, as you can see, my final answer does not match that of the text book. However, I've been told that my answer is correct, so long as the value expressed is a maximum. Can anyone help me out?

    Many thanks.


    Q.
    The given diagram (see attachment) shows part of the graph of y=x^2. p(x, y) is a point on the curve and a=(6,0). bp is perpendicular to ab. (i) Express the coordinates of p in terms of x only, (ii) Find the value of x, if the area of the \triangle abp is a maximum &, hence, find the maximum area.

    Attempt: (i) If y=x^2, then p(x,x^2)

    (ii) From the diagram, we can infer that b=(x,0).
    Area of \triangle abp: \frac{1}{2}|(y_2-y_1)(x_1-x_3)-(x_2-x_1)(y_1-y_3)|
    Subbing in the vertices of the triangle: \frac{1}{2}|(0-0)(6-x)-(x-6)(0-x^2)|=\frac{1}{2}|0-(-x^3+6x^2)|=\frac{1}{2}|x^3-6x^2|=
    \frac{x^3-6x^2}{2}
    Find derivative to determine value of x: \frac{dA}{dx}=\frac{1}{2}(3x^2-12x)=0\rightarrow3x^2-12x=0\rightarrow x(x-4)=0\rightarrow x= 0 or x=4
    Second derivative determines maximum/ minimum area: \frac{d^2A}{dx^2}=3x-6\rightarrow3(4)-6=6>0, where area is a minimum when x=4
    Area: \frac{x^3-6x^2}{2}=\frac{4^3-6(4)^2}{2}=\frac{64-96}{2}=\frac{-32}{2}=-16

    Ans.: (From text book): x = 2, Area = 8
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  2. #2
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    Re: Maximum and Minimum Problems

    Quote Originally Posted by GrigOrig99 View Post
    The question statement indicates that the area will be a maximum, but in my attempt you can see that I have arrived at a positive value of 6, suggesting a minimum. I think this has something to do with how I've applied the modulus when calculating the area of the triangle. Specifically, I 'probably' can't define \frac{1}{2}|x^3-6x^2| as either \frac{x^3-6x^2}{2} or \frac{-x^3+6x^2}{2}. Furthermore, as you can see, my final answer does not match that of the text book. However, I've been told that my answer is correct, so long as the value expressed is a maximum. Can anyone help me out?

    Many thanks.


    Q.
    The given diagram (see attachment) shows part of the graph of y=x^2. p(x, y) is a point on the curve and a=(6,0). bp is perpendicular to ab. (i) Express the coordinates of p in terms of x only, (ii) Find the value of x, if the area of the \triangle abp is a maximum &, hence, find the maximum area.

    Attempt: (i) If y=x^2, then p(x,x^2)

    (ii) From the diagram, we can infer that b=(x,0).
    Area of \triangle abp: \frac{1}{2}|(y_2-y_1)(x_1-x_3)-(x_2-x_1)(y_1-y_3)|
    Subbing in the vertices of the triangle: \frac{1}{2}|(0-0)(6-x)-(x-6)(0-x^2)|=\frac{1}{2}|0-(-x^3+6x^2)|=\frac{1}{2}|x^3-6x^2|=
    \frac{x^3-6x^2}{2}
    Find derivative to determine value of x: \frac{dA}{dx}=\frac{1}{2}(3x^2-12x)=0\rightarrow3x^2-12x=0\rightarrow x(x-4)=0\rightarrow x= 0 or x=4
    Second derivative determines maximum/ minimum area: \frac{d^2A}{dx^2}=3x-6\rightarrow3(4)-6=6>0, where area is a minimum when x=4
    Area: \frac{x^3-6x^2}{2}=\frac{4^3-6(4)^2}{2}=\frac{64-96}{2}=\frac{-32}{2}=-16

    Ans.: (From text book): x = 2, Area = 8
    If \displaystyle \begin{align*} b < 6 \end{align*}, the base of the triangle is \displaystyle \begin{align*} 6 - x \end{align*} and the height is \displaystyle \begin{align*} x^2 \end{align*}, so the area will be \displaystyle \begin{align*} A = \frac{1}{2}\left( 6 - x \right) x^2 = 3x^2 - \frac{1}{2}x^3 \end{align*}

    Differentiating gives \displaystyle \begin{align*} \frac{dA}{dx} = 6x - \frac{3}{2}x^2 \end{align*}. The critical points are where \displaystyle \begin{align*} \frac{dA}{dx} = 0 \end{align*}, so

    \displaystyle \begin{align*} 6x - \frac{3}{2}x^2 &= 0 \\ 12x - 3x^2 &= 0 \\ 3x \left( 4 - x \right) &= 0 \\ x = 0 \textrm{ or } x &= 4 \end{align*}.

    There's not much point checking the nature of x = 0, since this will give 0 area and is clearly a minimum. As for the other, the second derivative is \displaystyle \begin{align*} \frac{d^2A}{dx^2} = 6 - 3x = -6 \end{align*} when \displaystyle \begin{align*} x = 4 \end{align*}. Since the second derivative is negative at this critical point, it's a maximum.

    So the maximum area is \displaystyle \begin{align*} A = 3 ( 4 )^2 - \frac{1}{2} (4)^ 3 = 48 - 32 = 16 \end{align*}.

    Your book has a mistake.
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  3. #3
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    Re: Maximum and Minimum Problems

    Thank you very much.
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