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 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)

shubham garg Grade: 12

        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)

7 years ago

Answers : (1)

AJIT AskiitiansExpert-IITD

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 !!!

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation. 


Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.




Win exciting gifts by answering the questions on Discussion Forum..