# Calculus

posted by on .

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 - ,

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.