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Q) A window is in the shape of a rectangle surmounted by a semicircle,If the Perimetre of the Window is Pcm show that the Window lest in the maximum possible light only when the radius of the semicircle is P/(Ï€+4)cm.?
Prove it!!!!!!!
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10 Answers
62perimeter is given by
2H+B+Ï€B = P
H=P-(1+Pi)B/2
area is BH+Ï€B2/4
=B{ P-(1+Pi)B/2} + πB2/4
dA/dB=0
is it now obvious?
62See the max sunlight will come if the area is max
perimeter is fixed
base is given by B
there is a semicirle of diameter B
and a rectangle of base B and height H
find the area of it
elimnate either B or H wrt Perimeter.
that is all :)
1Perimeter P=2(x+y)+1/2* 2pi.r
where r=radius of the semicircle and x and y r the sides of rectangle.....
and y=2r... cant we do like that way bhaiyya...........
P=2x+4r+pi.r
= 2x+r(4+pi)...........rite bhaiyya.....
62yes we can do this way also terminator..
but what is the issue then? !!