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November 23, 2014

November 23, 2014

Posted by **math** on Monday, October 26, 2009 at 6:26pm.

- Calculus -
**MathMate**, Monday, October 26, 2009 at 7:32pmLet ABC be the vertices of the triangle, right-angled at B, AB=5, BC=12 (vertical side), AC=13.

Draw a rectangle BDEF, where D is on AB, E is on AC and F is on BC.

Denote

x=DE= height of rectangle

Width of rectangle = DB = 5-(5x/12)

Area of rectangle,

A(x)=x(5-(5x/12))=5x-5x²/12

A'(x) = 5-10x/12

For A(x) to be maximum,

A'(x) = 0 = 5-10x/12

x=6, 5-5(6)/12 = 2.5

The maximum area is 6*2.5=15

Note that the aspect ratio of the rectangle is the same as that of the right sides of the triangle.

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