Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #79

Find the greatest area of a rectangle with sides parallel to the axes and vertices on x²/a²+y²/b²=1, a,b>0.

✅ Step-by-Step Solution

Final Answer: 2ab; dimensions √2a by √2b.

Let its first-quadrant vertex be (u,v). Then v=b√(1−u²/a²), 0≤u≤a.
Its area is A=4uv≥0, so maximize A²=16b²(u²−u⁴/a²).
Differentiation gives (A²)′=32b²u(1−2u²/a²), with interior zero u=a/√2.
The derivative changes from positive to negative there; endpoint areas vanish. Thus v=b/√2 and the global maximum is attained.
The rectangle dimensions are √2a by √2b, and Amax=4(a/√2)(b/√2)=2ab.

[1 mark] Let its first-quadrant vertex be (u,v). Then v=b√(1−u²/a²), 0≤u≤a.
[1 mark] Its area is A=4uv≥0, so maximize A²=16b²(u²−u⁴/a²).
[1 mark] Differentiation gives (A²)′=32b²u(1−2u²/a²), with interior zero u=a/√2.
[1 mark] The derivative changes from positive to negative there; endpoint areas vanish. Thus v=b/√2 and the global maximum is attained.
[1 mark] The rectangle dimensions are √2a by √2b, and Amax=4(a/√2)(b/√2)=2ab.

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