Question

The maximum area of a rectangle whose two consecutive vertices lie on the x-axis and another two lie on the curve y = e^-|x|, is equal to

• A. 2e sq. units
• B. 2/e sq. units
• C. e sq. units
• D. 1/e units.

Answer 1

Answer: (B)

2/e sq. units

Answer 2

Let the rectangle be ABCD, with A ≡ (t, 0), B ≡ (t, e^-t). C ≡ (-t, e^−t), D ≡ (-t, 0) where

t ∈ R⁺. Area of rectangle ABCD = 2t. e^-t = f(t) (say)

Now = 2(t (−e^−t)+ e^-t) = 2e^-t (1 – t).

Clearly $\frac { \mathrm { df } } { \mathrm { dt } }$ > 0, t ∈ (0, 1) and < 0 for t ∈ (1, ∞) . Thus t = 1 is the point of maxima for f(t). Thus maximum area = f(1)

= $\frac { 2 } { e }$