Answer
$P(2-\sqrt 3),\dfrac{P(3-\sqrt 3)}{6},\dfrac{P(2\sqrt 3-3)}{3}$
Work Step by Step
From the given figure, we have $c= b \cos \theta$
Area of a pentagon is: $A=b \cos \theta (b \sin \theta +2a)$
Perimeter of a pentagon is: $P=2 (a+b+b \cos \theta)$
Need to apply Lagrange Multipliers Method to determine the dimensions of a rectangular box of maximum volume.
we have $\nabla f=\lambda \nabla g$
Using constraint condition, we get
$a=b(\dfrac{\sqrt 3+1}{2})$ and $\theta = \pi/6$
Thus, $P=2 (a+b+b \cos \theta) \implies b=\dfrac{P}{2\sqrt 3+3}$
The third side can be found as: $2c =2b \cos \theta$
$=2(\dfrac{P}{2\sqrt 3+3}) \cos (\pi/6)$
$=P(2-\sqrt3)$
Hence, Sides of a pentagon are: $P(2-\sqrt 3),\dfrac{P(3-\sqrt 3)}{6},\dfrac{P(2\sqrt 3-3)}{3}$
