Q.368. Maximising area of a trapezoid.

Question 368. Find the maximum area of a trapezoid whose area is given by
A = 4sin(x)(16+16cos(x)) = 64sin(x) + 64sin(x)cos(x).

Answer 368.
For A to be maximum, dA/dx = 0 and d^2A/dx^2 < 0
dA/dx = 0
=> 64cosx + 64cos2x = 0
=> cos2x + cosx = 0
=> 2cos^2 x + cosx - 1 = 0
=> 2cos^2 x + 2cosx - cosx - 1 = 0
=> (2cosx - 1) (cosx + 1) = 0
=> cosx = 1/2 or - 1
=> x = 60° or 270°
d^2A/dx^2 = - 64sinx - 128sin2x < 0 for x = 60°
and maximum area
=A (max)
= 64sin60° + 32sin120°
= 96sin60°
= 83.14 sq. units.

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