On Wed, 6 Feb 2008 16:10:25 -0800 (PST), baz90@[EMAIL PROTECTED]
wrote:
>Hello all,
> I have just done 35 calculus questions of varying difficulty
>and solved them quite easily, but the last problem i can't get my
>thinking around . This is it: A rectangle PQRS is placed inside
>a scalene triangle ABC [ the diagram is supplied but I can't draw it;
>to explain the diagram.. BC is the base of ABC; P is on AB;Q is on AC;
>S and R are on BC]. If the area of the triangle ABC is constant,
>prove that the maximum area of the rectangle is one-half the area of
>the triangle ABC.
> Hope somebody can even get me started. Thank you. bazza
Set your triangle up with its base on the x-axis. Say B = (a,0) and C
= (b, 0), with B to the left of C, and the third vertex on the y axis
at (0, h). You can do that by placing the third vertex on the y-axis
first and let a and b fall wherever they are; it doesn't matter if one
or both is negative.
Now show the line AB has the equation:
y = (- h/a )* (x -a)
and line AC has the equation
y = (- h/b )*(x - b)
Now draw a rectangle in the triangle with height m. The top side of
this rectangle will intersect lines AB and AC at points, call them V
and W, respectively. You know the y coordinates of V and W are m and
you can figure out the x coordinates by putting x = m in corresponding
equations above.
Show that the x coordinate of V is ( a - am/h) and the x coordinate of
W is (b - bm/h). Then show the length of the rectangle is x-right
minus x-left, which comes out (b - a)(1 - m/h)
So the area of the rectangle is:
A = length * height = (b - a)*(1 - m/h)*m
= (b - a)*(m - m^2/h), a quadratic equation in m.
Now, finally, you are done with the *algebra* ****tion of the problem
and can do the calculus problem. You have the area as a function of m.
Find its max using calculus and verify it is half the area of the
triangle. Of course, this last step could be done without calculus
too...
--Lynn
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