# Maximum minimum problem

1. Jul 23, 2011

1. The problem statement, all variables and given/known data
The base b and the area k of a triangle are fixed. Determine the base angles if the angle at the vertex opposite b is to be a maximum.

2. Relevant equations

3. The attempt at a solution
The answer the book gives is arctan(4k/b2)

I looked at the solution to the Regiomontanus' angle maximization problem but I could not modify some part of the solution to get this.

Also I tried to put everything in terms of h and b.(ie. if k and b are constant, h(height) also must be constant. So all the triangles we get with different angles are constructed if you put a base on a line and put a point on the paper such that when you connect the lines to the edge of the base to the point the required constant area should be achieved. Then just move the point in a line parallel to the base. Then becomes clear from a qualitative viewpoint(not rigorous) that the maximum angle should be there when the point intersects the altitude of the base. But still how do I get this base angle.(Use lagrange's multipliers to help or just traditional calculus)?

2. Jul 24, 2011

### nickalh

I've got a system of 4 equations with 5 variables. It sounds worse than it is, because none of the 5 are particularly difficult, compared to http://www.mathematics2.com/Calculus/RegiomontanusAngleMaximizationProblem". That can simplify into one equation with two variables. Be sure to keep the top angle, in my case, $\beta$, so we can maximize $\beta$.

Technical note: My system of equations assumes neither of the base angles are 90. I suspect the top angle is > 90, so will ignore this case for now.

Last edited by a moderator: Apr 26, 2017
3. Jul 24, 2011

### nickalh

Hints:
Write out which letters are constants.
Write out which letters are variables.

I don't use either side connecting with the top vertex. This implies I'm not using Pythagorean theorem either.

My version is based on Calculus I methods, with an esoteric derivative thrown in. Originally, I had arctangent, but I took deriv. of arccotangent, because it allowed me to put variables in the numerator. That maybe getting ahead of ourselves.

Write out as many equations as you think are relevant. Then pick which ones have the fewest variables & will simplify the most. Show me what all your equations you've got and how you've simplified them.

4. Jul 24, 2011

### Ray Vickson

So the base b and height h are fixed (with k = b*h/2). It is easiest to think of the base as the interval (-b/2,b/2) on the x-axis in an xy-plane, and the vertex as lying on the line y = h, at the point (x,h). The angle you want can be written as angle = f(x) + f(b-x) for some function f that you can figure out.

RGV

5. Jul 25, 2011

### nickalh

Isn't that assuming the optimal triangle is isosceles? From how I read the problem, the goal is to prove what you're assuming.

6. Jul 25, 2011

### SammyS

Staff Emeritus
Let's see if OP can now do some work on this problem -- or at least see if OP will respond.

7. Jul 25, 2011

### Ray Vickson

No. It holds for all x from -infinity to +infinity. however, I should have said that the base goes from 0 to b, not from -b/2 to b/2.

RGV

8. Jul 25, 2011

### Hariraumurthy

If x be the distance from one endpoint to the place the altitude intersects the base produced in both directions, and B be the angle opposite b, then the following equation holds.

B=arctan(x/h) + arctan[(b - x)/h] where h = 2k/b

This equation even holds if x is negative or larger than b

(take my figure that I have uploaded for the case of x is negative.(I will let you work you the case of x is larger than b). Theta=arctan[(b - x)/h](remember x is negative and this adds on to b). alpha=arctan(x/h). This will make alpha negative. B is just alpha + theta=arctan(x/h) + arctan[(b - x)/h]).)

If we take db/dx and set it to 0, we find that B is maximum when

b^2 - 2bx = 0, i.e. x=b/2

This proves that given the area and the base, the triangle with the largest angle opposite b is the triangle whose altitude bisects the base, i.e. an isosceles triangle with equal base angles.

This implies the tangent of both of the base angles is h/(b/2)=4k/(b^2)

Therefore both of the base angles = arctan[4k/(b^2)]

I hope I did not give away too much.

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9. Jul 26, 2011

### Ray Vickson

You gave away everything! Now the OP can get full marks without doing any of the required work.

RGV

10. Jul 26, 2011

### nickalh

Congratulations on finding a much simpler solution, than I did. Your equation with arctan is nearly identical to the one I simplified everything into.

The origin is at the point where x & b touch each other, I misread it the first time, for Hariraumurthy's answer.

In hindsight, discovered I should have asked the student to show the work he had done. Then we could have gone straight to the better methods.

For students to learn, they must actually do some work. The retention level goes up significantly when they at try, make some mistakes, ** then find or are told better ways. The post given does little to develop the student's ability to work this exercise, as challenging as it was.

There's a minor or possibly significant mistake in the result given. Keeping the previous paragraph in mind, I leave it to the original poster to find.

Last edited by a moderator: Apr 26, 2017
11. Jul 28, 2011

### Hariraumurthy

Actually, I am a vadiraja's brother and I know that he spent the required effort to try for the solution. He was really beating himself up about it that he could not find the solution to a problem of this nature. He gave ME an idea about the tan-1 and then I did not want to show the solution to him outright so I posted on here instead. He later actually found the solution independent of this and then I showed it to him. He is a very smart boy actually and did the solution for himself.

Last edited: Jul 28, 2011