Maximizing a Trapezoid

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In summary, the problem is to find the angle and dimensions that will maximize the carrying capacity of a gutter made from a long strip of sheet metal. The equation for the carrying capacity is given and the solution involves taking partial derivatives and setting them to 0. However, without a graphing calculator, the equation is difficult to solve and simplify. The process of factoring common terms can make it easier to solve, but it is important to check if any assumptions need to be made.
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jesuslovesu
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Homework Statement


A gutter is to be made from a long strip of sheet metal 24 cm wide, by bending up equal amounts at each side through equal angles. Find the angle and the dimensions that will make the carrying capacity of the gutter as large as possible.

Homework Equations


The Attempt at a Solution



The equation I came up with is A = 24lsin[tex]\theta[/tex] - [tex]2l^2sin\theta[/tex] + [tex].5*l^2sin(2\theta)[/tex]

And when I take the partial derivatives I get two equations and set them to 0
24sin[tex]\theta[/tex] - 4lsin[tex]\theta[/tex] + 2lsin[tex]\theta[/tex]cos[tex]\theta[/tex] = 0

24lcos[tex]\theta[/tex] - 2l^2cos[tex]\theta[/tex] + 2l^2cos[tex](2\theta)[/tex] = 0

Substitution:
[tex]\frac{24^2*sin(\theta)}{4sin\theta-sin(\theta*2)} - \frac{2*(24sin\theta)^2}{(4sin\theta-sin(2\theta))^2}cos\theta + \frac{2(24sin(\theta))^2*cos(2\theta)}{4sin\theta - sin(2\theta)^2}[/tex]

But how would one solve that without a graphing calculator? I can put it in and get 90 degrees (which I think is the correct answer). The book I'm using was written before calculators existed... I don't see any way to simplify that.
 
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jesuslovesu said:
Substitution:
[tex]\frac{24^2*sin(\theta)}{4sin\theta-sin(\theta*2)} - \frac{2*(24sin\theta)^2}{(4sin\theta-sin(2\theta))^2}cos\theta + \frac{2(24sin(\theta))^2*cos(2\theta)}{4sin\theta - sin(2\theta)^2}[/tex]

But how would one solve that without a graphing calculator? I can put it in and get 90 degrees (which I think is the correct answer). The book I'm using was written before calculators existed... I don't see any way to simplify that.

[tex]\frac{24^2*sin(\theta)}{4sin\theta-sin(2\theta)} - \frac{2*(24sin\theta)^2}{(4sin\theta-sin(2\theta))^2}cos\theta + \frac{2(24sin(\theta))^2*cos(2\theta)}{(4sin\theta - sin(2\theta))^2}=0[/tex]

Throw out some common factors and it's a bit easier to look at:
[tex]\sin(\theta) (4 \sin\theta - \sin(2 \theta))+ 2 \sin^2 \theta \cos \theta +2 \sin^2 \cos^2\theta[/tex]
(Warning - this assumes that [itex](4sin\theta - sin(2\theta))^2 \neq 0[/itex] -- you'll have to check whether that leads to an answer.)

Now, note that
[tex]2 \sin \theta \cos \theta = \sin 2\theta[/tex]

[tex]4 \sin^2 \theta - \sin^2 \theta \cos \theta + 2 \sin^2 \theta \cos \theta + 2\sin^2 \theta \cos^2 \theta=0[/tex]

Drop the [itex]\sin^2[/itex] since it's in all terms, and the rest is pretty straightforward.

P.S. You might want to follow this process to the end, and see what answer it leads to.
 

1) What is a trapezoid?

A trapezoid is a quadrilateral with one pair of parallel sides.

2) How do you find the area of a trapezoid?

The formula for finding the area of a trapezoid is A = 1/2 (a + b)h, where a and b are the lengths of the parallel sides and h is the height.

3) What is the difference between a trapezoid and a parallelogram?

A parallelogram has two pairs of parallel sides, while a trapezoid only has one pair of parallel sides.

4) Can you maximize the area of a trapezoid?

Yes, you can maximize the area of a trapezoid by increasing the length of the parallel sides and/or the height.

5) How is the perimeter of a trapezoid calculated?

The perimeter of a trapezoid is calculated by adding the lengths of all four sides together.

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