Maximizing Area of Cross-Section of Isosceles Trapezoid

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Homework Help Overview

The discussion revolves around maximizing the area of the cross-section of an isosceles trapezoid. Participants are exploring the relationships between the dimensions of the trapezoid and the area formula, as well as the necessary conditions for optimization.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to derive the area of the trapezoid using the formula and are considering the implications of maximizing this area. There are discussions about the need for derivatives and related equations, particularly concerning perimeter and other dimensions.

Discussion Status

Some participants are actively sharing their calculations and expressing confusion about the problem's requirements. Guidance has been offered regarding the differentiation of constants, indicating a productive exchange of ideas, though consensus on the approach has not been reached.

Contextual Notes

There is mention of constraints related to the learning material, specifically that derivatives of the sine function have not been covered, which may limit some participants' ability to progress in their solutions.

ha9981
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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.
 

Attachments

  • calctrapezoid.jpg
    calctrapezoid.jpg
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Hi ha9981! :wink:

Show us your full calculations, so we can see where the problem is, and we'll know how to help! :smile:
 
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!
 
ha9981 said:
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! :smile:

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

sin60 is a constant

differentiation not needed! :wink:
 

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