Eric Uwanyagasani kappy

This page lists questions and answers that were posted by visitors named Eric Uwanyagasani kappy.

Questions

The following questions were asked by visitors named Eric Uwanyagasani kappy.

Answers

The following answers were posted by visitors named Eric Uwanyagasani kappy.

let x be width and y be length ,and as we know perimeter of rectangle=2(length+widith) so 2(x+y)=100 x+y=50 , x=50-y area=length*width area=xy=(50-y)y area=50y-y^2 dA/dy=0 for min or max dA/dy=50-2x=0 y=25 x=25

let x be width and y be length ,and as we know perimeter of rectangle=2(length+width) so 2(x+y)=100 x+y=50 , x=50-y area=length*width area=xy=(50-y)y area=50y-y^2 dA/dy=0 for min or max dA/dy=50-2y=0 y=25 x=25