JEE Main 2021 (Online) 27th August Morning ShiftQuestion 1
A wire of length $20$ m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is :
Differentiation and its Application
Solution
Concept. Let the hexagon have side $s$ (using $6s$); the square uses the rest. Minimise the total area. Formulas used. (1) Area of a regular hexagon of side $s$: $\frac{3\sqrt3}{2}s^2$Step 1 - Area. The square has side $\frac{20-6s}{4}$, so $$A(s)=\frac{3\sqrt3}{2}s^2+\left(\frac{20-6s}{4}\right)^2.$$ Step 2 - Derivative. $$A'(s)=3\sqrt3\,s-\frac{3(20-6s)}{4}=0\ \Longrightarrow\ 4\sqrt3\,s=20-6s.$$ Step 3 - Solve. $s=\dfrac{20}{6+4\sqrt3}=\dfrac{10}{3+2\sqrt3}$, and $A''>0$, so it is the minimum. Answer: (C)