IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

nkatkamwar

it goes like this "Maximum area of rectangle with base on +ve X-axis and two vertices lie on a line y-x+3=0 and 2y+x-10=0.Find maximum area of rectangle?"
okay guys plz answera this ,its taking our brains out!!!!!

netkid07

so u want max area

right ?

rhd92781

is the ans 49/12.

i took AB as y=0, then CD is y=k. BC and AD are parallel to x=0

if D lies on y-x+3=0 and C on 2y+x-10=0

then D(k+3, k), C (10-2k, k), B (10-2k, 0) A (k+3, 0)

now area Ar=AB*AD

= (7-3k)*k

for max area, d(A)/dk =0

or 7-6k = 0 of k = 7/6

so max area = 49/12

please rate if correct

iitkgp_bipin

Well done rhd92781.

Take the length of the other side = k and the find the equation of the opposite side. Then this line and two given lines will give you the vertices. Now find area and maximize the result.

nkatkamwar

Thanks man its correct!!!!!

nadeemoidu

I'm not sure if I got the question rightly but I think the above method is wrong.

Firstly you have not made any use of the fact that the base lies on the +ive x axis , which is given in the question.

Now, why is it necessary that D lies on y-x+3=0 and that C lies on 2y+x-10=0?

Cant it be that C lies on y-x+3=0 and that D lies on 2y+x-10=0 ?
The area will then be (3k-7)*k , which does not have any maxima .
But when you take the condition that the base lies on the +ive x axis , then the maximum area can be found to be 40 sq units

nkatkamwar

But the ans is correct. Its the question from Arihant diff calculas but older version. i tried to find innew one but failed . the ans is 49/12, which is coming only by taking the given lines with given eq parellal. but slopes dont come equallll!! PLZ correct me if i m wrong and give right concept. plz

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