Asked by Sarskia

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?
Answered by MathMate
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)
Answered by Reiny
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 ....
Answered by Reiny
go with Damon

I skimmed over the "parabolic section" of your post.
Answered by Sarskia
..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.
Answered by MathMate
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

Answered by Sarskia
Okay, thank you very much!
Answered by MathMate
You're welcome!
Answered by Sarskia
I had a run over it myself and got the maxima as (-2,-3) same numbers, but negative form?
Answered by MathMate
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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