Question

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​

Answer 1

Solution ........ .

Answer 2

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