optimization problem!

[johnjohnsen]

Hi guys! I am not sure if I posted this in the right section,

Picture: http://gyazo.com/277af69ef04be03187abba03fbb1f617

I got a problem with optimization of a triangle.
The length is 300
(its a 3d shaped object.)
the length of AD is 15 and CD 15. AC is open and can be adjusted.

My question is, at which degree of B is the volume the biggest.

Thanks already,

Im not sure how to make the solution,

because I don't know how to calculate the 2 unknown (height and width)

 

[Ishuda]

Hi guys! I am not sure if I posted this in the right section,

Picture: http://gyazo.com/277af69ef04be03187abba03fbb1f617

I got a problem with optimization of a triangle.
The length is 300
(its a 3d shaped object.)
the length of AD is 15 and CD 15. AC is open and can be adjusted.

My question is, at which degree of B is the volume the biggest.

Thanks already,

Im not sure how to make the solution,

because I don't know how to calculate the 2 unknown (height and width)

Assuming AC is parallel to the horizontal axis, draw a line from D to AC. That line will be perpendicular to AC and the two triangles will be congruent. Let the bottom angle be x [the one inside the triangle] that will give you w and h as a function of x. Maximize the area which will maximize the volumn.

got to go for now.

 

[johnjohnsen]

Assuming AC is parallel to the horizontal axis, draw a line from D to AC. That line will be perpendicular to AC and the two triangles will be congruent. Let the bottom angle be x [the one inside the triangle] that will give you w and h as a function of x. Maximize the area which will maximize the volumn.

got to go for now.

Yes i know that, but im having a problem making the formula of the maximized volume.

I know i need to replace B with x.

But how do i get the x with the size of the line that is parallel to AC.

I only know AD! im thinking about something with: volume = 360(sin(x)) x something else...

 

[HallsofIvy]

The volume is the length, 300, times the area of the triangle. And the area of the triangle, as a function of angle B, can be can be determined by the "cosine law". If each outside angle has measure, B, then the angle in the triangle, at vertex D, has measure 180- 2B and the opposite side has length c, with \(\displaystyle c^2= 15^2+ 15^2- 2(15)(15)cos(180- 2B)= 225(1+ 2cos(2B))\). The altitude can be calculated by dropping a perpendicular from point D to the other side, dividing it into two right angles with angle at D of measure 90- D and "hypotenuse" 15. So this triangle has base \(\displaystyle 15\sqrt{1+ 2cos(2B)}\) and altitude \(\displaystyle 15 cos(90- B)\).

Use that to write the area as a function of B. Take the derivative and set it equal to 0 to find the B that maximizes the area and so volume.

 

[Ishuda]

Yes i know that, but im having a problem making the formula of the maximized volume.

I know i need to replace B with x.

But how do i get the x with the size of the line that is parallel to AC.

I only know AD! im thinking about something with: volume = 360(sin(x)) x something else...

Slightly different way: The two triangles formed have the angle x at the bottom of the triangle, i.e.
2x = \(\displaystyle \angle\)ADC
h = a cos(x)
w = 2 a sin(x)
where a is the length of DC, w is the length of AC and h is the height of triangle ACD. Thus the area A of triangle ACD is given by
A(x) = h * (w/2) = a² sin(x) cos(x) = 0.5 a² sin(2x)

Is that (or HallsofIvy's post) sufficient?

 

[stapel]

Picture: http://gyazo.com/277af69ef04be03187abba03fbb1f617

I got a problem with optimization of a triangle.
The length is 300
(its a 3d shaped object.)
the length of AD is 15 and CD 15. AC is open and can be adjusted.

Just to be clear: Are you saying that the shape is a right triangular prism, so that what you've drawn is the cross-section? And that, in order to maximize the volume of the entire shape, you need to maximize the area of the displayed triangle?

If so, where are you stuck in applying the formula for the area A of a triangle with base b (in this case, the length of AC) and height h (in this case, the length of the perpendicular from the midpoint M of AC down to D)? Having drawn the height line, having noted the right triangles this forms (such as AMD), having noted that the measure of angle B must be the same as the measure of angle MAD, and having applied geometric or trigonometric info, what have you done? Where are you stuck?

Please be complete. Thank you!

 

[johnjohnsen]

The volume is the length, 300, times the area of the triangle. And the area of the triangle, as a function of angle B, can be can be determined by the "cosine law". If each outside angle has measure, B, then the angle in the triangle, at vertex D, has measure 180- 2B and the opposite side has length c, with \(\displaystyle c^2= 15^2+ 15^2- 2(15)(15)cos(180- 2B)= 225(1+ 2cos(2B))\). The altitude can be calculated by dropping a perpendicular from point D to the other side, dividing it into two right angles with angle at D of measure 90- D and "hypotenuse" 15. So this triangle has base \(\displaystyle 15\sqrt{1+ 2cos(2B)}\) and altitude \(\displaystyle 15 cos(90- B)\).

Use that to write the area as a function of B. Take the derivative and set it equal to 0 to find the B that maximizes the area and so volume.

o that

so, that means that the volume is calculated by: \(\displaystyle 300 * 15\sqrt{1+ 2cos(2B)} * 15 cos(90- B)\). than I would differentiate that, and i would set it to 0. Right?