Posted by Sarskia on .
The curved section can be modelled by the parabola
y=3x3/4x^2
Use calculus to find the maximum height of the curved parabolic section.
How do I do this step by step?

Calculus 
MathMate,
If you have done functions before calculus as most students do, you can rewrite the equation in the standard form:
f(x)=a(xh)+k
where the vertex is found at
(h,k)
or
f(x)=(3/4)(x2)²+3
where the vertex is found at (2,3).
Confirm by Calculus:
f(x)=3x3/4x^2
f'(x)=3(3/2)x
At the vertex, f'(x)=0
x=3*(2/3)=2
y=3(2)(3/4)(2²)
=63
=3
So the vertex is at (2,3) 
Calculus 
Reiny,
your equation is ambiguous
do you mean
y = (3x3)/(4x^2) ? or
y = 3x  3/(4x^2) ? or
y = ((3x3)/4)(x^2) or
y = 3x  (3/4)x^2 ? or .... 
Calculus 
Reiny,
go with Damon
I skimmed over the "parabolic section" of your post. 
Calculus 
Sarskia,
..The last one.
Confirm by Calculus:
f(x)=3x3/4x^2
f'(x)=3(3/2)x
At the vertex, f'(x)=0
x=3*(2/3)=2
y=3(2)(3/4)(2²)
=63
=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 
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 
Calculus 
Sarskia,
Okay, thank you very much!

Calculus 
MathMate,
You're welcome!

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

Calculus 
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.