Step-by-Step Solution

Topic: Application of Derivatives (Maxima & Minima)

(I) RELATION BETWEEN x AND y

Dimensions given: Length = 2x, Breadth = 2y.

Since the semi-circle surmounts the top length (2x), the diameter is 2x. Thus, Radius r = x.

Perimeter (P) = 10m

The perimeter of the window includes the bottom base, two vertical sides, and the semi-circular arc (the top side of the rectangle is not included in the outer boundary).

P = Base + 2(Height) + Semi-circle Arc

10 = 2x + 2(2y) + πr

10 = 2x + 4y + πx

4y = 10 – 2x – πx

(II) AREA FUNCTION A(x)

Total Area = Area of Rectangle + Area of Semi-circle

A = (Length × Breadth) + (1/2)πr²

A = (2x)(2y) + (1/2)πx² = 4xy + 0.5πx²

Substitute 4y from Part (i): 4y = 10 – x(2 + π)

A(x) = x(4y) + 0.5πx²

A(x) = x[10 – 2x – πx] + 0.5πx²

A(x) = 10x – 2x² – πx² + 0.5πx²

(III)(A) MAXIMIZE AREA

Differentiate A(x) w.r.t x:

A'(x) = 10 – 4x – πx = 10 – x(4 + π)

Set A'(x) = 0 for critical points:

10 = x(4 + π)   ⇒ x = 10 / (4 + π)

Find y:

4y = 10 – x(2 + π). Substitute x:

4y = 10 – [10(2 + π) / (4 + π)]

4y = [10(4 + π) – 10(2 + π)] / (4 + π)

4y = [40 + 10π – 20 – 10π] / (4 + π) = 20 / (4 + π)

y = 5 / (4 + π)

(III)(B) ALTERNATIVE VARIABLES (x=Length, y=Breadth)

If Length = x and Breadth = y:

Radius r = x/2.

Perimeter Relation: x + 2y + π(x/2) = 10

2y = 10 – x – (πx/2)

Area A = xy + (1/2)π(x/2)²

A = x(y) + πx²/8

Substitute y = [10 – x – πx/2] / 2:

A(x) = (x/2)[10 – x – πx/2] + πx²/8