1] A window is in the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is of constant length k then its maximum area is
Answers
Solution ........ .
Toolbox:
Area of a circle =πr2
Area of a rectangle =lb
d\{d}{x} ( x^{n} ) = nx^{n-1}
Step 1:
Perimeter of the window when the width of window is x and 2r is the length.
⇒2x+2r+12×2πr=10[Given]
2x+2r+πr=10
2x+r(2+π)=10------(1)
For admitting the maximum light through the opening the area of the window must be maximum.
A=Sum of areas of rectangle and semi-circle.
Step 2:
Area of circle=πr2
Area of rectangle=l×b=2×r×x
A=2rx+12πr2
=r[10−(π+2)r]+12πr2
=10r−(12π+2)r2
For maximum area dAdr=0 and d2Adr2 is -ve.
⇒10−(π+4)r=0
(π+4)r=10
r=10π+4
Step 3:
d2Adr2=−(π+4)[Differentiating with respect to r]
(i.e)d2Adr2 is -ve for r=10π+4
⇒A is maximum.
From (1) we have
⇒10=(π+2)r+2x
Put the value of r in (1)
10=(π+2)×(10π+4)+2x
10=10(π+2)π+4+2x
=10(π+2)+2x(π+4)π+4
10(π+4)=10(π+2)+2x(π+4)
10(π+4)−10(π+2)=2x(π+4)
10π+40−10π−20=2x(π+4)
20=2x(π+4)
10=x(π+4)
x=10π+4
Step 4:
Length of rectangle=2r=2(10π+4)
=20π+4
breadth=10π+4