maximizing the amount of water in a viaduct....

[bluester105]

Hey, thanks for taking the time to read this! I'm just super confused and not really understanding what my teacher is telling me. if yall would please help me out it would be greatly appreciated. thanks soo much!

A=(1/2)*h*(b1+b2) where b1 and b2 are the two parallel bases and h is the altitude.

Your problem: a water viaduct has a cross section that is an isosceles trapezoid. the area of the trapezoid varies as angle theta changes. you want to determine the angle theta that will MAXIMIZE the amount of water this viaduct will conduct.

…...\.................|........................../
…….\...............|........................./
……...\8…….h…………......8/
……….\...........|……………/
….(theta\_____|_________/theta)
............................8.............
i have no idea how the 8 FEET got there just that, this is the diagram he gave us and that was on it.

First: determine a trigonometric expression for the height--find the height in terms of angle theta.

this is what i put..im not sure if this is right... soo
A=(1/2) h (b1+b2)
2A=2(1/2)h(b1+b2)
2A=h(b1+b2)
H=2A/(b1+b2)

Find an expression for the length of the upper base in terms of angle theta

again not sure if this is right..

2A=h(b1+b2)
b1+b2= (2A)/h
b2=-b1 + (2A)/h

from here on out I have no idea what my teacher is asking, if yall would please help me in the right direction that would be soo helpful. thanks...
Find an expression for the area of the viaduct in terms of angle theta.

use third grade algebra to simplify your expression for the area-i.e. combine terms; factor our common factors etc.

now use calculus to find the angle that will MAXIMIZE the area of the viaduct, and thus maximize the volume of water flowing through it.

 

[wjm11]

[FONT=&quot]2A=h(b1+b2)
H=2A/(b1+b2)

[/FONT]Find an expression for the length of the upper base in terms of angle theta

[FONT=&quot]again not sure if this is right..

2A=h(b1+b2)
b1+b2= (2A)/h
b2=-b1 + (2A)/h

from here on out I have no idea what my teacher is asking, if yall would please help me in the right direction that would be soo helpful. thanks...
[/FONT]Find an expression for the area of the viaduct in terms of angle theta.

Hello, Bluestar,

You’ve been ignoring the “in terms of theta” instructions.

Height: sin(theta) = opp./hyp. = h/8; h = 8[sin(theta)]

Upper base: Recognizing that, for logical reasons, theta must be less than 90 degrees, b2 (top base) equals b1 (bottom base) plus an amount dependent on theta.

b2 = 8 + 2[8cos(theta)]

You already have the equation for the area of the trapezoidal aqueduct. Now just substitute in the b1 and b2 that are written in terms of theta.

Make sense?

After that, take the derivative with respect to theta and find the zeroes (to locate maxima or minima).

 

[bluester105]

thanks for the help!

so for the area of the viaduct i got

A=(1/2) (8 sin (theta)) (8+8+16cos(theta))
A=(1/2) (8 sin(theta)) (16+16cos (theta))
A=64 sin(theta) (1+cos(theta))

soo to find the angle that will maximize the area of the viaduct and the volume of water i did...

A'=64 [sin(theta)(-sin(theta)) + (1+cos(theta))(cos(theta))]
A'=64 [-sin^2(theta)+cos^2(theta) + cos(theta)=0
[cos^2(theta)-sin^2(theta)+cos(theta)]=0

is this where i use trig identities? i think this goes into a quadradic equation but the minus sign is throwing me off...