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Thread: Max/Min Application

  1. #1
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    Exclamation Max/Min Application



    The parabola y = x^2 and the line y = mx +b intersect at the points A (α, α^2) and B (β, β^2) as shown.

    a) Explain why αβ = - b

    b) Show that AB = √ (m^2 + 4b) (m^2 + 1)

    c) The point P(x, x^2) lies on the parabola between A and B. Show that the area of triangle APB is given by:

    Area of Triangle APB = (b + mx x^2) √ (m^2 + 4b)

    d) Find the position of point P that maximizes the area of triangle APB in terms of m.


    Could someone please help me solve this problem?
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post


    The parabola y = x^2 and the line y = mx +b intersect at the points A (α, α^2) and B (β, β^2) as shown.

    a) Explain why αβ = - b

    b) Show that AB = √ (m^2 + 4b) (m^2 + 1)

    c) The point P(x, x^2) lies on the parabola between A and B. Show that the area of triangle APB is given by:

    Area of Triangle APB = (b + mx x^2) √ (m^2 + 4b)

    d) Find the position of point P that maximizes the area of triangle APB in terms of m.


    Could someone please help me solve this problem?


    to #a):
    Use the 2-point-formula of a straight line:

    $\displaystyle \dfrac{y-\alpha^2}{x-\alpha}=\dfrac{\beta^2-\alpha^2}{\beta-\alpha}$

    $\displaystyle y-\alpha^2=\dfrac{(\beta^2-\alpha^2)(x-\alpha)}{\beta-\alpha}$

    $\displaystyle y=\dfrac{((\beta-\alpha)(\beta+\alpha))(x-\alpha)}{\beta-\alpha}+\alpha^2$

    $\displaystyle y=(\beta+\alpha)x-\alpha \beta-\alpha^2+\alpha^2$

    $\displaystyle y=(\beta+\alpha)x-\alpha \beta$ Compare this equation with

    $\displaystyle y=m \cdot x+b$ which gives $\displaystyle b = - \alpha \beta$
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  3. #3
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    Quote Originally Posted by xwrathbringerx View Post


    The parabola y = x^2 and the line y = mx +b intersect at the points A (α, α^2) and B (β, β^2) as shown.
    ...
    d) Find the position of point P that maximizes the area of triangle APB in terms of m.


    Could someone please help me solve this problem?
    The area of the triangle is a function with respect to x. Calculate the first derivation and set this term equal to zerO.

    $\displaystyle A(x)=\dfrac12(b+mx-x^2)\underbrace{\sqrt{m^2+4b}}_{constant!}$

    $\displaystyle A'(x)=\dfrac12 \sqrt{m^2+4b} \ \cdot \ (m-2x)$

    $\displaystyle A'(x) = 0~\implies~m-2x=0~\implies~x=\dfrac m2$
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  4. #4
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    Sorry but how do I find the area of the triangle APB?
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  5. #5
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    Quote Originally Posted by xwrathbringerx View Post


    The parabola y = x^2 and the line y = mx +b intersect at the points A (α, α^2) and B (β, β^2) as shown.

    a) Explain why αβ = - b

    b) Show that AB = √ (m^2 + 4b) (m^2 + 1)
    b) From part a) you got, $\displaystyle \alpha\beta=-b$

    Also, slope of given line (y=mx+b) = slope of AB

    $\displaystyle m = \frac{\beta^2-\alpha^2}{\beta-\alpha}=\beta+\alpha$

    Now, use distance formula to calculate distance $\displaystyle AB= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

    $\displaystyle AB=\sqrt{(\beta-\alpha)^2+(\beta^2-\alpha^2)^2}$

    $\displaystyle AB=\sqrt{(\beta-\alpha)^2+(\beta-\alpha)^2(\beta+\alpha)^2}$

    $\displaystyle AB=\sqrt{(\beta-\alpha)^2+(\beta-\alpha)^2m^2}$

    $\displaystyle AB=\sqrt{(\beta-\alpha)^2(1+m^2)}$

    Now,

    $\displaystyle (\beta-\alpha)^2 = \beta^2-2\alpha\beta+\beta^2$

    $\displaystyle = \beta^2+2\alpha\beta+\beta^2-4\alpha\beta$

    $\displaystyle =(\beta+\alpha)^2-4\alpha\beta$

    $\displaystyle =m^2-4(-b)=m^2+4b$

    $\displaystyle (\beta-\alpha)^2=m^2+4b$


    Therefore,

    $\displaystyle AB=\sqrt{(\beta-\alpha)^2(1+m^2)} = \sqrt{(m^2+4b)(1+m^2)}$
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    Quote Originally Posted by xwrathbringerx View Post


    The parabola y = x^2 and the line y = mx +b intersect at the points A (α, α^2) and B (β, β^2) as shown.

    c) The point P(x, x^2) lies on the parabola between A and B. Show that the area of triangle APB is given by:

    Area of Triangle APB = (b + mx – x^2) √ (m^2 + 4b)
    AB is the base of triangle, and perpendicular distance from P to AB is height of triangle.

    Height = Perpendicular distance from $\displaystyle P(x, x^2)$ to line mx + b - y = 0

    $\displaystyle = \frac{|mx + b - x^2|}{\sqrt{1^2+m^2}}$

    $\displaystyle = \frac{|mx + b - x^2|}{\sqrt{1+m^2}}$

    Area of triangle $\displaystyle = \frac{1}{2}\times base \times height$

    $\displaystyle = \frac{1}{2}\times AB \times height$

    $\displaystyle =\frac{1}{2}\times \sqrt{(m^2+4b)(1+m^2)} \times \frac{|mx + b - x^2|}{\sqrt{1+m^2}}$

    $\displaystyle =\frac{1}{2}(mx + b - x^2) \sqrt{(m^2+4b)} $

    $\displaystyle Area,\;\;A(x)=\frac{1}{2}(mx + b - x^2) \sqrt{(m^2+4b)} $
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  7. #7
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    Quote Originally Posted by xwrathbringerx View Post

    d) Find the position of point P that maximizes the area of triangle APB in terms of m.

    d) The area of the triangle is a function with respect to x. Calculate the first derivation and set this term equal to zero

    $\displaystyle Area,\;\;A(x)=\frac{1}{2}(mx + b - x^2) \sqrt{(m^2+4b)} $

    $\displaystyle A(x)=\dfrac12(mx+b-x^2)\underbrace{\sqrt{m^2+4b}}_{constant!}$

    $\displaystyle A'(x)=\dfrac12 \sqrt{m^2+4b} \ \cdot \ (m-2x)$

    $\displaystyle A'(x) = 0~\implies~m-2x=0~\implies~x=\dfrac m2$

    $\displaystyle
    Point \;\;P(x, x^2)=\left(\frac{m}{2}, \frac{m^2}{4}\right)$

    Did you get all the parts NOW ???
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  8. #8
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    *scratches head*

    How exactly does become ?

    Is m - 2x the factorised form of mx + b - x^2?
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  9. #9
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    Quote Originally Posted by xwrathbringerx View Post
    *scratches head*

    How exactly does become ?

    Is m - 2x the factorised form of mx + b - x^2?
    I have calculated the first derivation of A. Since a constant factor stays unchanged I have marked the constant factor, meaning: Don't touch me!

    Therefore you only have to derive $\displaystyle (mx+b-x^2)$ with respect to x (the derivation of b wrt x is zero):

    $\displaystyle \dfrac{d(mx+b-x^2)}{dx }= m-2x$
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