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Calculus

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The curved section can be modelled by the parabola
y=3x-3/4x^2
Use calculus to find the maximum height of the curved parabolic section.
How do I do this step by step?

  • Calculus - ,

    If you have done functions before calculus as most students do, you can rewrite the equation in the standard form:
    f(x)=a(x-h)+k
    where the vertex is found at
    (h,k)
    or
    f(x)=-(3/4)(x-2)²+3
    where the vertex is found at (2,3).

    Confirm by Calculus:
    f(x)=3x-3/4x^2
    f'(x)=3-(3/2)x
    At the vertex, f'(x)=0
    x=3*(2/3)=2
    y=3(2)-(3/4)(2²)
    =6-3
    =3
    So the vertex is at (2,3)

  • Calculus - ,

    your equation is ambiguous

    do you mean
    y = (3x-3)/(4x^2) ? or
    y = 3x - 3/(4x^2) ? or
    y = ((3x-3)/4)(x^2) or
    y = 3x - (3/4)x^2 ? or ....

  • Calculus - ,

    go with Damon

    I skimmed over the "parabolic section" of your post.

  • Calculus - ,

    ..The last one.

    Confirm by Calculus:
    f(x)=3x-3/4x^2
    f'(x)=3-(3/2)x
    At the vertex, f'(x)=0
    x=3*(2/3)=2
    y=3(2)-(3/4)(2²)
    =6-3
    =3
    So the vertex is at (2,3)

    A question regarding the first step (differentiation) How do you differentiate the fractions?
    I'm stuck.. And clearly thick. Haha.

  • Calculus - ,

    You do not really differentiate the fractional numbers, they are just the coefficients which just "stick around".
    f(x)=ax²+bx
    f'(x)=2ax+b
    So in the above,
    a=-(3/4), b=3
    f'(x)=2(-3/4)x+3
    =-(3/2)x+3

  • Calculus - ,

    Okay, thank you very much!

  • Calculus - ,

    You're welcome!

  • Calculus - ,

    I had a run over it myself and got the maxima as (-2,-3) same numbers, but negative form?

  • Calculus - ,

    Check your work, draw a graphical sketch to convince yourself that there is only one maximum.

    Also, f(2)=3, and f(-2)=-9, so (-2,-3) does not lie on the curve.

    If all fails, post your working for a check.

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