Of all the rectangles each of which has a perimeter 40 m find which one had maximum area.find the area also.

Let l and b be the length and breadth of rectangle and A is its area.perimeter=2l+b40=2l+b20=l+bb=20-l     .....1Now,A=lbA=l20-lA=20l-l2Differentiating both sides w.r.t. l, we getdAdl=20-2lFor the maximum area, its derivative will be zero.dAdl=020-2l=02l=20l=10 mPutting l=10 m in 1, we getb=20-10=10 mWhen length and breadth both are equal to 10 m, area will be maximum.Area=lb=10×10=100 m2

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Area will be maximum when both L,B=10. but the it is a square then.so,the one with L=11 and b=9 wil be rectangle with the maximum area=99. You can also say when L=10.1and B=9.9 is the rectangle with maximum area=99.99.
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