1. maximizing volume

the problem is, A rain gutter is to be made from a metal sheet of width 36cm by bending up 1/4th of the sheet on both sides through angle theta. how should theta be chosen so that the gutter can carry the maximum amount of water?

I'm having trouble getting an equation for this problem any ideas?

2. Originally Posted by sparowe
the problem is, A rain gutter is to be made from a metal sheet of width 36cm by bending up 1/4th of the sheet on both sides through angle theta. how should theta be chosen so that the gutter can carry the maximum amount of water?
maximize cross-sectional area of the trapezoid ...

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

where ...

$\displaystyle h = 9\sin{\theta}$

$\displaystyle b_1 = 18$

$\displaystyle b_2 = 18 + 2(9\cos{\theta})$

find $\displaystyle \frac{dA}{d\theta}$ and maximize.

3. I'm having trouble getting A' i got 324cos(theta)+81cos^2(theta)-81sin^2(theta) and i'm not quiet sure how to get the critical points of it.

4. Originally Posted by sparowe
I'm having trouble getting A' i got 324cos(theta)+81cos^2(theta)-81sin^2(theta) and i'm not quiet sure how to get the critical points of it.
correction ...

$\displaystyle A' = 162\cos{\theta}+81\cos^2{\theta}-81\sin^2{\theta} = 0$

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

now use the identity

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

to get the expression quadratic in $\displaystyle \cos{\theta}$ ... you'll need to use the quadratic formula to solve.

5. after doing the quadratic i got, (-1+/-sqrt(12cos^2(theta))/2

i'm not 100% sure how to put cos^2 x into the calculator i did (cos(x))^2

anyway i tried to do the first derivative test on the the exact answer of the quadratic and got two positives so i messed up somewhere

6. Originally Posted by sparowe
after doing the quadratic i got, (-1+/-sqrt(12cos^2(theta))/2

i'm not 100% sure how to put cos^2 x into the calculator i did (cos(x))^2

anyway i tried to do the first derivative test on the the exact answer of the quadratic and got two positives so i messed up somewhere

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

$\displaystyle a = 2$ , $\displaystyle b = 2$ , $\displaystyle c = - 1$

$\displaystyle \displaystyle \cos{\theta} = \frac{-2 \pm \sqrt{2^2 - 4(2)(-1)}}{2(2)}$

you will get two values on the right hand side of the above equation ... one is valid, one is not. Use the inverse cosine function to find the angle $\displaystyle \theta$ that maximizes the area.

7. I'm getting 1.232 as the maximum, but when i plug this into cos^-1 it says error on my calculator, did i miss a step some where?

8. Originally Posted by sparowe
I'm getting 1.232 as the maximum, but when i plug this into cos^-1 it says error on my calculator, did i miss a step some where?
$\displaystyle \cos{\theta} = \frac{-2 \pm \sqrt{12}}{4} = \frac{-1 \pm \sqrt{3}}{2}$

$\displaystyle \frac{-1 + \sqrt{3}}{2} = 0.366025...$

$\displaystyle \frac{-1 - \sqrt{3}}{2} = -1.366025...$

the domain of the inverse cosine function is between -1 and 1 ... as I told you previously, one of those solutions is invalid.

using the valid solution, $\displaystyle \theta = \arccos(0.366025...) \approx 68.5^\circ$

you need to see your instructor.

9. i know, when i simplified the problem (-2+/-sqrt(12))/4 i did not reduce the 2 in 2sqrt(3)