What is the area of the largest triangle that can be fitted into a rectangle of length 10 m and width 9 m ?

Let OA=10cm along X axis and OB be 9 cm along Y axis. Let OACB be the rectangle.

The triangle fitted in OACB has to have two of its vertices on any two adjacent sides, say at X  On OA and  Y on OB and the third vertex on any of the other 2 sides. But if you make the other vertex of triangle coincide with C, then the triangle XYC fitted in the rectangle OACB will be maximum for a given  position of X on OA and Y on OB.

Now  the area of this triangle XYC is (by coordinate geometry) = (1/2){OA*OY+OB*OX- OX*OY}

=(1/2){l*y+bx-xy}, where l = OA and b = OB, OX =x and OY =y

Therefore area of XYC will be maximum when partial derivatives wrt x and Y are zero and d2/dxdy of area A(x,y) of triangle XYC is -ve. So,

d/dx A(x,y) = 0= d/dy A(x,y) and  for these x and y ,d2/dxdy A(x,y) is -ve.

d/dx A(x,y) = 0 gives b-y = 0 and d/dy A = 0 gives l-x = 0

So x =l and y=b and

Obviously (d2/dxdy A) at x=l, y=b is  -1.

Therefore, the maximum area of the triangle XYC = 1/2{lb+lb-lb} = (1/2)lb

=(1/2)10*9 = 45 sq units Or half the area of the rectangle.