Let `A(p^2,-p),B(q^2, q),C(r^2,-r)` be the vertices of triangle ABC. A parallelogram AFDE is drawn with D,E, and F on the line segments BC, CA and AB, respectively. Using calculus, show that the maximum area of such a parallelogram is `1/2(p+q)(q+r)(p-r)`.

Explanation

From similar triangle s FBD and ABC , `(c-x)/(c )=(y)/(b)`
or `y=(b//c)c-x`
`therefore` Z=Area of AFDE= `xy sin A = b (sin A)/(c ) (cx-x^(2))`
`0ltxltc`
`therefore (dz)/(dx)=(b sin A)/(c )(c-2x)=0 pr x=c//2`
`(d^(2)Z)/(dx^(2))_(x=c//2)=(-2)/(c ) bsinAlt0`
Thus Z has maxima at `x= (c )/(2)`.So the greatest area of parallelogram AFDE is
`(b/c)msin A (c^(2)//a)=(1)/(2)((1)/(2)bc sin A)`
`=(1)/(2)triangle_(ABC)=(1/2)xx||{:(p^(2),,-p,=,1),(q^(2),,q,=,1),(r^(2),,-r,=,1):}||`
`=(1)/(4)||{:(p^(2),,-p,=,1),(q^(2)-p^(2),,q+p,=,0),(r^(2)-p^(2),,-r+p,=,1):}||R_(2)rarrR_(2)-R_(1)
R_(3)rarrR_(3)-R_(1)`
`=(1)/(4)|(q+p)(q+r)(p-r)|`