SOLUTION: find the dimentions of a rectangle( a )with the greatest area whose perimeter is 30 feet.

Question 49270: find the dimentions of a rectangle( a )with the greatest area whose perimeter is 30 feet.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
find the dimentions of a rectangle( a )with the greatest area whose perimeter is 30 feet.
PERIMETER(L+B)2=30
L+B=15
AREA = LB=L(15-L)
A=-(L^2-15L)=-[L^2-2(L)(7.5)+7.5^2-7.5^2]=7.5^2-(L-7.5)^2
SINCE (L-7.5)^2 IS ALWAYS GREATER THAN EQUAL TO ZERO,AREA IS MAXIMUM WHEN THIS IS ZERO
HENCE WHEN L-7.5=0....OR...L=7.5,AREA IS MAXIMIUM AND EQUAL TO 7.5^2
HENCE L=B=7.5 FOR MAXIMUM AREA
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