Question

A window is in the form a rectangle surmounted by a semicircular opening the total perimeter of the window is 10m.find the dimensions of the window to admit maximum light through the whole opening. plz concept main batna​

Answer 1

Q)

A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

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A)


Step 1:

Perimeter of the window when the width of window is xx and 2r2r is the length.

⇒2x+2r+12⇒2x+2r+12×2πr×2πr=10=10[Given]

2x+2r+πr=102x+2r+πr=10

2x+r(2+π)=102x+r(2+π)=10------(1)

For admitting the maximum light through the opening the area of the window must be maximum.

A=A=Sum of areas of rectangle and semi-circle.

Step 2:

Area of circle=πr2πr2

Area of rectangle=l×b=2×r×xl×b=2×r×x

A=2rx+12πr2A=2rx+12πr2

=r[10−(π+2)r]+12πr2=r[10−(π+2)r]+12πr2

=10r−(12=10r−(12π+2)r2π+2)r2

For maximum area dAdrdAdr=0=0 and d2Adr2d2Adr2is -ve.

⇒10−(π+4)r=0⇒10−(π+4)r=0

(π+4)r=10(π+4)r=10

r=10π+4r=10π+4

Step 3:

d2Adr2d2Adr2=−(π+4)=−(π+4)[Differentiating with respect to r]

(i.e)d2Adr2d2Adr2 is -ve for r=10π+4r=10π+4

⇒A⇒A is maximum.

From (1) we have

⇒10=(π+2)r+2x⇒10=(π+2)r+2x

Put the value of rr in (1)

10=(π+2)×(10π+4)10=(π+2)×(10π+4)+2x+2x

10=10(π+2)π+410=10(π+2)π+4+2x+2x

=10(π+2)+2x(π+4)π+4=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)

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)10π+40−10π−20=2x(π+4)

20=2x(π+4)20=2x(π+4)

10=x(π+4)10=x(π+4)

x=10π+4x=10π+4

Step 4:

Length of rectangle=2r=2(10π+4)=2r=2(10π+4)

=20π+4=20π+4

breadth=10π+4