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shubham garg Grade: 12
        

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

Answers : (1)

AJIT AskiitiansExpert-IITD
68 Points
										

Dear  Shubham , 1124_21440_Untitled.gif


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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