Minimize triangle area
A right triangle, shown below,the right angle at origin. What are the dimensions of x and y such that the hypotenuse passes thru point (2,5) and the triangle area is minimized.
Another Image (click)
Labels: mathschallenge, SharedPuzzle
posted by Ragknot at
7:11 AM
15 Comments:
x=4, y=10
is this your homework?
lol
if this is your homework, gj
pretty basic optimization problem
No, it's not my homework, but many years ago it might have been an assignment I made to students.
Now it should be a challenge to many who care to accept it. Too many nowdays can't think of any way to solve it.
this is really a challenging problem.
Ragknot,
you made a mistake. the picture is ABOVE the words, not below them. :)
Oh, and I also meant to say that I agree with you, Rajesh. This is a tough question.
:)
:)
~Bre
First reply above is right.
If the "top" of the triangle on the y axis is, say, at 5+n then we find that the rightmost point on the x axis is at 2+10/n.
So the area xy/2 = (5+n)(2+10/n)/2
or 10+n+25/n
This has a minimum where n = 5,
i.e. y = 10 and x = 4.
The previous was right at
to the "Area = 10n+25/n."
Next could have been the derivative of that would be
A(n)=1-25/n^2
0=1-25/n^2
1=25/n^2
and solving... n = 5
x = 5
y = 8
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Those are very very very pretty pictures. But, they are very boring.
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