Maths Question, Can Anyone Help?

Question

A V-Shaped animal trough which is made up of two planks of wood having width 0.5 metre and length 6 meters closed at the end by ends by a vertical sheet of timber. Determine the depth of the trough such that it holds a maximum amount of water?

The cross section is V- Shaped.

I know i have to set up an equation, differentiate it find maximums bla bla bla. But i cant set up the Equation!

Can anyone help?

Im not given any angles so how can trig help?

Answer 1

The required depth = 0.3536m, or about 35.4cm

This can be solved with or without reference to the angle between the trough sides. The volume V = 6 A where A is the area of a triangular end. So we just have to find when A is a maximum.

A = b*h/2 (where h is the required depth)

Also, b = 2*sqrt(0.5^2 - h^2)

so A = h*sqrt(0.5^2 -h^2)

Now find h, such that: dA/dh =0

or, more simply:

The angle betwen the 2 pieces which maximizes the volume is 90 degrees(see footnote). Once you know this you can determine h since h = 0.5*cos(45) = 0.3526

Footnote: Express the the end area in terms of the angle between the 2 pieces of wood. Use the area formula: A = (1/2)a*b*sin(x) where x is the angle between the two pieces. In your case, a=b= 0.5m and so A = 0.5*0.5*sin(x)/2 = 0.125sin(x).
So the volume V= 6*0.125*sinx(x) = 0.75sin(x) cubic m. This will be a maximum when x = 90 degrees since we know the sine function value is a maximum when x=90 degrees, but formally:

dV/dx = 0.75 d/dx sin(x) = 0.75cos(x)=0 when cos(x) =0 when x = 90degrees
Max volume = 0.75sin(90) = 0.75 m^3

Answer 2

The two sides of the trough will be 0.5m with an unknown angle between them. The vertical line from the bottom of the V to the surface is the unknown - call this 'd'.
I think the best way is to find the angle that maximises the volume.
The depth d forms the adjacent to a right angled triangle with the angle being half of what you are trying to find. The 0.5m is the hypotenuse. Try to draw what I've described. The depth d is equal to 0.5Cos(angle/2). The water surface forms the third side of the triangles. Half of the water surface is the side of the right angled triangle, call it w and so:
w =0.5Sin(angle /2).

The formula for the area of one right angled triangle is half length x width, so the area of the V is twice this and equal to w x d.

Putting the angles into the equation:
Area = 0.5Cos(angle/2) x 0.5Sin(angle/2).

I think you need to differentiate and look for the zeros, one will be the maximum. It's about 25 years since I differentiated cos and sin, so I can't help you there. Once you've found the angle that maximises the area, then put it back into d = 0.5Cos(angle/2) to find the depth.

I hope I've not confused you too much - it certainly made me use a few brain cell.

Answer 3

There's one little problem which hasn't been addressed. In order to make the trough, one of the planks has to be nailed to the other, so its effective width is no longer 0.5 meter, but 0.5 - thickness of the plank. If the planks are 5 cm thick, for example, the width of one of the planks is reduced to 0.45 meters. My guess is that a standard right-angle join will result in a triangular cross-section with the largest area, hence the greatest trough capacity. Draw a sketch with the top of the trough as the base/hypotenuse of a right-angled triangle formed by the two planks as viewed on end, allowing for the reduced width of one of them. Now drop a perpendicular from the apex of the triangle to the base, and figure out how to calculate that, which will be the depth of the trough. I think you're going to need some trigonometry to get the base angles. If you can draw an accurate scale diagram, you can get a pretty good estimate of the exact measurement. Good luck.

Answer 4

Everyone is making such a big deal of this, it is not hard.

First, since the volume is the length times the surface area of the triangular base, and the length is a constant 6 meters, all we care about maximizing is the area of the triangle.

Draw a triangle with 2 sides .5 and angle between them 2a, then draw the altitude with angle a. The height of the triangle is .5cos(a) and half the base is .5sin(a), so the area of the triangle is .25sin(a)cos(a) = .125sin(2a). This is clearly maximized when sin(2a) = 1 (no calculus even necessary), so the guesses about a right angle are right.

Answer 5

Since the length is fixed at 6 meters, it suffices to find the maximum area of a triangle where two of the sides are known to equal 0.5 meters, because the volume of the trough is just the area of the triangle times 6. (I suspect it's going to be an equilateral triangle.) I'll let you sweat the details.

Answer 6

forget the length. Just make the angle such that the area of the triangle is maximized. With the angle very small or very large, the area is about zero. It's easier if you draw a vertical line and deal with 1/2 the triangle (now looking at a right triangle). So find the max area of a triangle with a constant hypotenuse. (also forget the 1/2) a2 x b2 = c2 a = sq rt of c2 - b2 b = sq rt of c2 - a2
So, area is (sq rt of c2 - b2) x (sq rt of c2 - a2)
hint - the answer is about 0,35 metre

Answer 7

i think you need to find the third side of a triangle with 2 sides known (each 0.5 m) such that the area is maximum.
the length of the thing doesnot matter.
Since an equilateral is the solution,now find the height of this triangle.

Answer 8

It's when area of cross-section (of triangle) is max.
Area of triangle= 1/8(2sinacosa)*
A = 1/8(sin2a)
dA/da =1/4(cos2a)
max/min when cos2a=pi/2
or a=pi/4=45deg
> d= 0.5sin 45
> d= 0.3536m for depth
* area triangle = 1/2x(base)x(height)

Answer 9

More information required.