If the sum of three dimensions and the total surface area of a rectangular box are 12 cm and 94 cm² respectively, then the maximum length of a stick that can be placed inside the box is

Let length = $$l$$, breadth = b, height = h
Given that
($$l$$ + b + h) = 12 cm
Total surface area of box
= 2($$l$$b + bh + h$$l$$)
= 94 m² (Given)
⇒ ($$l$$ + b + h)² = $$l$$² + b² + h² + 2($$l$$b + bh + h$$l$$)
⇒ (12)² = $$l$$² + b² + h² + 94
⇒ 144 - 94 = $$l$$² + b² + h²
⇒ 50 = $$l$$² + b² + h²
Diagonal of box $$ = \sqrt {{l^2} + {b^2} + {h^2}} $$
∴ Length of longest rod that can be put inside the box
$$\eqalign{ & = \sqrt {{l^2} + {b^2} + {h^2}} \cr & = \sqrt {50} \cr & = 5\sqrt 2 {\text{ cm}} \cr} $$