Finding maximum cross-sectional area of isosceles trapezoid inscribed in semicircle

[ha9981]

Homework Statement

attached. It is just part a i am attempting for now.

Homework Equations

area of trapezoid = h(b1+b2 / 2)
where h is height, b1 is base one and b2 is base two.

The Attempt at a Solution

i really tried, i didn't know where to start. the answer i got was 30cm for base, while the correct answer was 20cm for base and sides. I am not sure where to go with this problem my first try was a guess and obviously didn't work properly. I think maybe i need someone to better explain this problem to me.

 

[tiny-tim]

Hi ha9981!

Show us your full calculations, so we can see where the problem is, and we'll know how to help!

 

[ha9981]

I am stuck on understanding exactly what the problem is asking.
I know i have to maximize the area, which i have the equation for, so ill find the derivative and set it to zero (due to turning points which are max and mins).

But I need a related equation as well, is it going to be perimeter?


here is my attempt:

Let x be the length folded.
Let y be the width folded.

total length = 5 - 2x

total width = AD + BC + AB
0.60 = 2y + AB

also i know:

x = BCsin60
x = ysin60

area of trapezoid = ysin60 ((0.60+(0.60-2y)) / 2)
area of trapezoid = ysin60 ((1.20 - 2y)/2)

Now i know there is something fatally wrong here since we didn't learn derivatives of sin function so i can't progress!

 

[tiny-tim]

area of trapezoid = ysin60 ((0.60+(0.60-2y)) / 2)
area of trapezoid = ysin60 ((1.20 - 2y)/2)

Now i know there is something fatally wrong here since we didn't learn derivatives of sin function so i can't progress!

Hi ha9981!

(i haven't checked your formula, but …)

sin60 is a constant …

differentiation not needed! ​