a trapezoid is a quadrilateral with (only) two sides (called bases) that are parallel. an isosceles trapezoid has two equal legs. the area of a trapezoid is
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.
attempt at what it looks like:
\ | /
\81 h 81/
\ | /
(theta\___|_____/theta)
8
first determine a trigonometric expression for the height-find the height in terms of angle theta.
find an expression for the length of the upper base in terms of angle theta.
find an expression for the area of the viaduct in terms of angle theta.
use 3rd grade alebra to simplify the 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.
The diagram didn't make it. I can figure that theta is the angle the side makes with the horizontal, but where does the 81 come in?
dang it. well the diagram was what my teacher gave me to go by. the 81 is the diagonals on both sides, or that's the way its written on the sheet of paper. and 8 is the length of the base.
To determine the angle theta that will maximize the amount of water the viaduct will conduct, we can follow these steps:
Step 1: Find a trigonometric expression for the height of the viaduct in terms of angle theta.
Since the cross-section is an isosceles trapezoid, the height (h) is the perpendicular distance between the two parallel bases.
From the given diagram, we can see that the height is opposite to angle theta. Therefore, using trigonometry, we can express the height as h = 8*sin(theta).
Step 2: Find an expression for the length of the upper base in terms of angle theta.
From the given diagram, the sum of the lengths of both bases is 81 units. Let's call the length of the upper base x, so the length of the lower base would be (81 - x) units since they are parallel.
Since the viaduct is isosceles, the two legs are equal. Therefore, the length of the upper base can be expressed as x = (81 - (81 - x))*tan(theta).
Simplifying this expression, we get x = 81*tan(theta) - x*tan(theta).
Bringing x*tan(theta) to the left side, we get x + x*tan(theta) = 81*tan(theta).
Factoring out x, we can rewrite it as x(1 + tan(theta)) = 81*tan(theta).
Dividing both sides by (1 + tan(theta)), we get x = (81*tan(theta))/(1 + tan(theta)).
Step 3: Find an expression for the area of the viaduct in terms of angle theta.
The area of a trapezoid is given by A = (1/2)*h*(b1 + b2), where h is the height and b1, b2 are the lengths of the parallel bases.
Using the expressions from Step 1 and Step 2, we can substitute these values into the area formula:
A = (1/2)*(8*sin(theta))*((81*tan(theta))/(1 + tan(theta)) + (81 - (81*tan(theta))/(1 + tan(theta))).
Step 4: Simplify the expression for the area using algebra.
To simplify this expression, we need to combine like terms and factor out common factors.
A = (1/2)*(8*sin(theta))*((81 + 81*tan(theta) - 81*tan(theta))/(1 + tan(theta))).
Simplifying further, we get A = (1/2)*(8*sin(theta))*(81/(1 + tan(theta))).
A = 4*sin(theta)*(81/(1 + tan(theta))).
Step 5: Use calculus to find the angle that maximizes the area of the viaduct.
To find the angle theta that maximizes the area, we need to find the derivative of the area function with respect to theta (dA/d(theta)) and set it equal to zero.
Taking the derivative, we get dA/d(theta) = 4*cos(theta)*(81/(1 + tan(theta))) - 4*sin(theta)*(81*tan(theta)/(1 + tan(theta))^2).
Setting dA/d(theta) = 0 and solving for theta will give us the angle that maximizes the area.
Due to the complexity of the expression, the final step requires a numerical approach or graphing calculator to find the specific value of theta that maximizes the area.
Note: It is assumed that the width of the viaduct is constant and doesn't vary with theta in this problem.