Integration- Finding minimum shaded value where x is p

You should know the area of the parabola, so subtract off the area of the triangle (similar to part a). What do you get? Then think about when that value is a minimum.

Alternatively, think about when the triangle has maximum area so when you subtract it from the parabola ...

I think I’m getting confused with the x’s and p’s

I found the area of the parabola Is 4/3a^3
I think I’m confused with the calculation for the area of the triangle.

When you google area of a triangle, what does it say?

I know right angle triangle is 1/2(h x w) it’s the calculating both parts of the triangle and then subtracting that from the parabola that’s currently giving me a headache. It’s been a long day for me. I’ll come back to it with a fresh head tomorrow
It’s been a long day

That formula works (base * perpendicular height)/2 even when its not right angled.
Both parts (base, perpendiular height) are relatively simple as is the result which maximises the area of the triangle.

Morning,

I’ve calculated these unknowns based on the information that I have. So, Q is equivalent to the bottom value.


I’ve calculated the area between p and a then deducted the triangle from that. If my working out makes sense. I just don’t know what I am looking in terms of the found answer.
If my approach is poor, then let me know.

Il check through your algebra now, but you're doing an approach which cause you problems.

The parabola has area 4/3a^3
The whole triangle has base 2a and height p^2, so its area is ...
The max area of the triangle is ..
Subtract max triangle from parabola ...

Each ... should be a single line (expression)

Taking the approach that you put across.

Wait, height p²?? The graph has a function y=a²-x²
So, the height of triangle is a²-p²? Have I lost my brain or something.

the base is the width on the x-axis, the height is y(p). I corrected the p^2 typo, in the previous post.

Lol. Okay .

This time it is actually the proper Archimedes parabola problem
https://www.math.mcgill.ca/rags/JAC/NYB/exhaustion2.pdf
One of the classic forerunners of integration.

Ooo, so it’s a famous question lol.

I’m looking and looking and the value of p doesn’t pop out to me lol.

Can I make the assumption the answer is ¹/₃a³ that’s the only familiarity between the two parts lol

The height of the triangle is max at p=0, so so is the area.
Or the parabola-triangle area is minimized when p^2 is minimum so p=0
Or ........

Note in your previous algebra heavy approach you dropped the /2 when subtracting the triangle from the parabola. And that was just half of the working I think.

Ooo, when I dropped the 2 I didn’t calculate the left side. The adjust the answer still didn’t help me.

I’m sure I have covered the triangle being max at 0 but I may have forgotten. Do you know what section that would be under?

There is "no working" for the triangle area being maximum - its simple gcse geometry, but you could use calculus - differentiation if you really wanted to. The area is
base*height/2
base = 2a which is constant
height = a^2 - p^2. Trivially, the maximum is a^2 which occurs at p=0.
So the maximum triangle area is at p=0 and is a^3.

Tbh you could do it your algebra heavy way, but you could simply deduce the answer by saying the parabola has constant area and the triangle has maximum area at p=0 (because it has the maximum height). So the answer is that worked out in part a).

GCSE wasn’t my strong point in life lol.

Although, I believe it just clicked now. The explanation as to why it’s peak is at 0. If we use an equilateral triangle as an example. On an x and y axis the y peak is where x=0.

This would be the same concept at peak of the scalene triangle.

Sort of, but its simply the triangle with the maximum height. Every other triangle (p not 0, with the same base) with a peak on the parabola has a smaller height and therefore a smaller area. Thats it.

The shape of the triangle is irrelevant, you just need the base and perpendicular height which corresponds to the x-y axes here.