Calculus
posted by math on .
What is the area of the largest rectangle that can be placed in a 51213 right triangle (as shown)?

Let ABC be the vertices of the triangle, rightangled 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))=5x5x²/12
A'(x) = 510x/12
For A(x) to be maximum,
A'(x) = 0 = 510x/12
x=6, 55(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.