Let A(p^2,-p), B(q²,q), C(r²,r) be the vertices of triangle ABC. a parallelogram AFDE is drawn with vertices D,E and F on the line segments BC,CA and AB respectively.Using calculus show that maximum area of such a parallelogram is

1/4(p+q)(q+r)(p-r)

Dear  Shubham ,

Please refer to the figure above.

area of parallelogram AFDE  =  base * length of altitude

                                                 = AF *AE sint =   hksint

Area of triangle ECD  =  1/2*(EC)*ED = 1/2*(m-k)*h*sint

Area of triangle FDB =  1/2*(FB)*FD = 1/2*(l-h)*k*sint

Summing the area to area of triangle ABC

(1/2)*lmsint =hksint +(1/2)*(m-k)*h*sint +(1/2)*(l-h)*k*sint

(1/2)*lm = hk +(1/2)*(m-k)*h+(1/2)*(l-h)*k

using equations we get , h = (lm - lk/m)

therefore area of AFDE  = (lm - lk/m)*k sint

to maximize this area differentiate this area w.r.t to k

d(area)/dt  = 0 will give k = m/2 and so h = l /2 .

therefore the points chosen are the mid points of the triangle.

therefore area of parallelogram is half the area of triangle .

therefore first find the area of traingle by matrix method and then half it ,

you will get the result .

GOOD luck !!!

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