
Re: Trigonometric area optimization
Posted:
Dec 15, 2012 8:54 AM


Ok, Adam. I'll give it a try, since you have furnished us with this very entertaining and really simple problem.
First, graph the line whose equation we have been given, which is y = 10  2x. We immediately know that 10 is the y intercept, and that the slope is 2, meaning it will decrease for increasing x. The x intercept can be found by setting y = 0, and we find that it is x = 5. So we have a picture in our mind, a line drawn between the points (0,10) and (5,0). Now we have the other equations, y = mx and y = 1/m x. These two lines meet at the origin of our coordinate system, and make a 90 degree angle with each other, as is evident by the relationship of their slopes. So we visualize how the right triangle consisting of the line segments formed by each of the legs of a right triangle emanating from the origin and intersecting our line y = 10  2x will appear.
It becomes apparent that a right triangle is formed, with the two sides subtending a segment from the line y = 10  2x serving as the hypoteneus.
Now as we rotate the lines a right angles forming the legs of the right triangle, we can imagine that, when the right triangle formed is one in which the legs are of equal length, this might indeed be the triangle that has the minimum area of all the other possible triangles. So we prove the following theorem, which is, that of all the right triangles formed by the setup, the isoceles one is the one that creates the minimum area.
So draw a right triangle meeting an hypotenus, and allowing the altitude that connects the vertex at the right triangle to the base to remain constant. Call one leg a and the other leg b. Call the angle made by a with the hypotenus, "h", then the angle that b makes with the hypotenus will be 90 degrees  h. The altitude will be a sine h. The base of the triangle will be a cos h + b cos (90  h) = a cos h + b sin h. So the whole area of the triangle will be 1/2 a sin h (a cos h + b sin h). We have one more relationship, which is that a sin h = b cos h, or a tan h = b. So we can replace b in our equation above. This gives us 1/2 a sin h (a cos h + a tan h cos h), or 1/2 a sin h (a cos h + a sin h). If we differentiate this with respect to the angle h, we get: (cos h  sin h)^2, and setting this equal to zero, we get cos h = sin h, which occurs only when h = 45 degrees.
In other words, if we draw a line perpendicular to the line y = 10  2x and passing through the origin, that line will form the altitude of a right triangle, whose sides are made by the two other lines y = mx and y = 1/m x. Now since you know the slope of the line, you also know the angle it makes with the x axis. So you should be able to find the angles the legs of the right triangle make with the x axis as well, and thus their slopes. Since I do not have a table of tangents right in front of me, I will let you do that. I diagram would have been very handy in this instance, wouldn't it have?

