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A=[xv(25-x2)]/2
dA/dx=(1/2).[(25-2x2)/ v(25-x2)]
put dA/dx=0 ,then we get
x=5/v2
Now d2A/dx2=(1/2)[(2x3-75x)/(25-x2)3/2]
Put x=5/v2, we get
D2A/dx2=-2 <0 so area is max at x=5/v2
So y2= 25-25/2=25/2
Y=5/v2
Thanks & Regards
Rinkoo Gupta
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