Step-by-Step Solution

Topic: Application of Derivatives

(I) AREA FUNCTION A(x)

From the ellipse equation x²/a² + y²/b² = 1:

y²/b² = 1 – x²/a² = (a² – x²)/a²

Taking square root: y = (b/a)√(a² – x²)

Area of Rectangle:

Area = Length × Breadth = (2x)(2y) = 4xy

Substitute value of y:

(II) CRITICAL POINT

To make differentiation easier, let us maximize Z = [A(x)]².

Z = 16x² (b²/a²) (a² – x²)

Z = (16b²/a²) (a²x² – x⁴)

Differentiate Z w.r.t x:

dZ/dx = (16b²/a²) (2a²x – 4x³)

Set derivative to zero:

2a²x – 4x³ = 0

2x(a² – 2x²) = 0

Since x ≠ 0 (dimension cannot be zero), we get:

2x² = a²   ⇒   x² = a²/2

(III) MAXIMIZATION & DIMENSIONS

Check for Maxima (Second Derivative Test):

d²Z/dx² = (16b²/a²) (2a² – 12x²)

At x = a/√2 (so x² = a²/2):

Value = K (2a² – 12(a²/2)) = K (2a² – 6a²) = -4Ka²

Since the result is negative, the area is MAXIMUM at this point.

Calculate Dimensions:

• Length (2x): 2(a/√2) = √2a
• Breadth (2y): First find y.
  y = (b/a)√(a² – a²/2) = (b/a)√(a²/2) = (b/a)(a/√2) = b/√2.
  Breadth = 2y = 2(b/√2) = √2b