On Feb 6, 6:10 pm, ba...@[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
Here's a solution simple enough to work out in your head. Suppose the
triangle has base (BC) of length b, and height 1.
Let w(x) be the width of the rectangle with height x, 0 <= x <= 1. It
should be obvious that w is a linear function, with w(0) = b and w(1)
= 0. So w(x) = b(1 - x). Now we want to maximize the area a(x) =
bx(1-x). Ignore the constant b and maximize x(1-x), then compute the
area of the rectangle.
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