# Optimization Question

• May 1st 2010, 05:57 PM
Paymemoney
Optimization Question
Hi
I need help on the following question:
1) A symmetrical gutter is made of metal sheeting 30cm. wide by bending it twice as shown in the cross section. Find an expression for the area $\displaystyle A(\theta)$ of the cross section and hence find the value of $\displaystyle \theta$ for the gutter to have maximum carrying capacity.

http://img94.imageshack.us/img94/2024/diagramit.jpg

This is what i have try to do so far.
I know that Area of a square is L*W and Area of 2 triangles is $\displaystyle 2(0.5*x^2 * sin(\theta))$

so $\displaystyle L*W + x^2 * sin(\theta) = 30$

Make $\displaystyle W = \frac{30 - x^2(sin(\theta))}{10}$

$\displaystyle A = L*W$

$\displaystyle A = 10(\frac{30 - x^2(sin(\theta))}{10})$

$\displaystyle A = 30 - x^2(sin(\theta))$

$\displaystyle A' = -x^2(\frac{dA}{d\theta})cos(\theta) + 2xsin(\theta)$

$\displaystyle A' = \frac{2xsin(\theta)}{-x^2(cos(\theta))}$

Don't think this is correct.
P.S
• May 1st 2010, 06:14 PM
skeeter
Quote:

Originally Posted by Paymemoney
Hi
I need help on the following question:
1) A symmetrical gutter is made of metal sheeting 30cm. wide by bending it twice as shown in the cross section. Find an expression for the area $\displaystyle A(\theta)$ of the cross section and hence find the value of $\displaystyle \theta$ for the gutter to have maximum carrying capacity.

http://img94.imageshack.us/img94/2024/diagramit.jpg

area of a trapezoid ...

$\displaystyle A = \frac{h}{2}(b_1+b_2)$

$\displaystyle h = 10\cos{\theta}$

$\displaystyle b_1 = 10$

$\displaystyle b_2 = 10 + 20\sin{\theta}$

try again.
• May 2nd 2010, 03:52 AM
HallsofIvy
Quote:

Originally Posted by Paymemoney
Hi
I need help on the following question:
1) A symmetrical gutter is made of metal sheeting 30cm. wide by bending it twice as shown in the cross section. Find an expression for the area $\displaystyle A(\theta)$ of the cross section and hence find the value of $\displaystyle \theta$ for the gutter to have maximum carrying capacity.

http://img94.imageshack.us/img94/2024/diagramit.jpg

This is what i have try to do so far.
I know that Area of a square is L*W and Area of 2 triangles is $\displaystyle 2(0.5*x^2 * sin(\theta))$

This makes no sense because you don't say what "x" means. Your area should depend on the single variable $\displaystyle \theta$, not both $\displaystyle \theta$ and some undefined "x".

What you can do is this: you know that the hypotenuse of each right triangle is 10 so, since "$\displaystyle sin(\theta)= \frac{opposite side}{hypotenuse}$", $\displaystyle opposite side= hypotenuse* sin(\theta)= 10sin(\theta)$ and "$\displaystyle cos(\theta)= \frac{near side}{hypotenuse}$", $\displaystyle near side= hypotenuse* cos(\theta)= 10 cos(\theta)$.

Of course, the area of a right triangle is "(1/2) near side* opposite side" so $\displaystyle A= \frac{1}{2}(10 cos(\theta))(10 sin(\theta))= 50 sin(\theta)cos(\theta)$.

The base of the central rectangle is 10 and the height is the "near side" of the two right triangles, $\displaystyle 10 cos(\theta)$ so its area is $\displaystyle 100 cos(\theta)$.

The entire area is $\displaystyle 100 cos(\theta)+ 100 sin(\theta)cos(\theta)= 100 cos(\theta)(1+ sin(\theta))$.

Quote:

so $\displaystyle L*W + x^2 * sin(\theta) = 30$

Make $\displaystyle W = \frac{30 - x^2(sin(\theta))}{10}$

$\displaystyle A = L*W$

$\displaystyle A = 10(\frac{30 - x^2(sin(\theta))}{10})$

$\displaystyle A = 30 - x^2(sin(\theta))$

$\displaystyle A' = -x^2(\frac{dA}{d\theta})cos(\theta) + 2xsin(\theta)$

$\displaystyle A' = \frac{2xsin(\theta)}{-x^2(cos(\theta))}$

Don't think this is correct.
P.S
• May 2nd 2010, 04:42 PM
Paymemoney
so then to find the value of $\displaystyle \theta$ do i make the equation:
$\displaystyle 100 cos(\theta)(1+sin(\theta)) = 30$
• May 2nd 2010, 05:40 PM
skeeter
Quote:

Originally Posted by Paymemoney
so then to find the value of $\displaystyle \theta$ do i make the equation:
$\displaystyle 100 cos(\theta)(1+sin(\theta)) = 30$

why are you setting the area = 30 ???

you want to find the value of $\displaystyle \theta$ that maximizes the cross-sectional area of the gutter ... i.e. the trapezoid.

find $\displaystyle \frac{dA}{d\theta}$ and maximize like you were taught.
• May 2nd 2010, 10:41 PM
Paymemoney
ok this is what i have done

$\displaystyle \frac{d(100cos(\theta)(1+sin(\theta)) = 900)}{dx}$

$\displaystyle 100cos(\theta)cos(\theta)+(1+sin(\theta))-100sin(\theta) = 0$

$\displaystyle 100cos^2(\theta) -100sin^2(\theta)-100sin(\theta) = 0$

$\displaystyle 1-2sin^2(\theta) - 100sin(\theta) =0$ this is where i get stuck, what should i do next?
• May 3rd 2010, 05:11 AM
skeeter
$\displaystyle A = 100\cos{\theta}(1 + \sin{\theta})$

using the product rule ...

$\displaystyle \frac{dA}{d\theta} = 100\cos{\theta}(\cos{\theta}) + (1+\sin{\theta})(-100\sin{\theta})$

$\displaystyle \frac{dA}{d\theta} = 100\cos^2{\theta} - 100\sin{\theta} - 100\sin^2{\theta}$

$\displaystyle 100\cos^2{\theta} - 100\sin{\theta} - 100\sin^2{\theta} = 0$

$\displaystyle (1-\sin^2{\theta}) - \sin{\theta} - \sin^2{\theta} = 0$

$\displaystyle 1 - \sin{\theta} - 2\sin^2{\theta} = 0$

$\displaystyle (1 - 2\sin{\theta})(1 + \sin{\theta}) = 0$

finish it