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```        Let  A(p2,-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) ```
7 years ago

68 Points
```										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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```
7 years ago
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