A covered metal triangular trough is constructed as follows:
A square shaped sheet of metal which is 70 centimeters wide and long square is folded along the center. Next, two pieces of metal in the shape of isosceles triangles are are welded to the ends. Finally, a metal cover is attached to the top).
We want to find the smallest and largest surfaces area S, that a so constructed trough can have, and at what opening angle w of the triangular pieces it is attained. Proceed as follows:
need to know:
Find the surfaces area as a function of the angle. (Be sure to include all five sides of the trough).
-- S(w) = ?
-- natural domain of S(w) (left and right end point)
On its domain S(w) has one stationary point, w=c. Although there is no explicit formula for the value of c, itself, it is possible to derive the exact value of the cosine of c,. Find it:
-- cos(c) =
Find the surface area at the stationary point
-- S(c) =
Find also the global minimal value of S in its natural domain
The surface area will be 70^2 + ends + top.
End surface area will be dependent on the angle:
altitude/35=sinTheta
area= 1/2 35*altitude= 1/2 *35^2*sinTheta
so two ends...35^2 sinTheta.
Now the top: 35*d where d is found by the law of sines: d/sinTheta=35/sin[1/2(180-theta)]
area top= 35*sintheat*35/sin(90-theta/2)
total area= sum of above.
To find the surface area as a function of the angle, we can substitute the given formulas for the end surface area and top surface area into the expression for the total surface area.
Let's denote the angle as θ.
The end surface area is given by:
End surface area = 2 * 35^2 * sin(θ)
The top surface area is given by:
Top surface area = 35 * sin(θ) * 35 / sin(90 - θ/2)
The total surface area is the sum of the square sheet area, the end surface area, and the top surface area:
S(θ) = 70^2 + 2 * 35^2 * sin(θ) + 35 * sin(θ) * 35 / sin(90 - θ/2)
Now, let's find the natural domain of S(θ), which means we need to determine the range of values for θ that make sense for the problem.
Since the triangular trough is formed by folding a square sheet along the center and attaching triangular pieces at the ends, the triangular pieces cannot extend beyond the edges of the folded square. This means that the opening angle, θ, should be between 0 and 180 degrees, or 0 and π radians.
Therefore, the natural domain of S(θ) is 0 ≤ θ ≤ π.
Next, we are asked to find the stationary point of S(θ) and the cosine of the angle corresponding to that stationary point.
To find the stationary point, we need to find the value of θ where the derivative of S(θ) is equal to zero.
By taking the derivative of S(θ) with respect to θ and setting it equal to zero, we can find the stationary point.
Finally, we can find the value of S(θ) at the stationary point, S(c), and determine the global minimum value of S(θ) within its natural domain by evaluating S(θ) at the endpoints of the domain and comparing it to S(c).