JEE Main 2022 (Online) 29th June Morning ShiftQuestion 1
A wire of length $22$ m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is :
Differentiation and its Application
Solution
Concept. Let the triangle have side $a$ (using length $3a$); the rest $22-3a$ makes the square. Minimise the total area in $a$. Formulas used. (1) Area of an equilateral triangle of side $a$: $\frac{\sqrt3}{4}a^2$Step 1 - Total area. The square has side $\frac{22-3a}{4}$, so by (1) $$A(a)=\left(\frac{22-3a}{4}\right)^2+\frac{\sqrt3}{4}a^2.$$ Step 2 - Derivative. $$A'(a)=-\frac{3(22-3a)}{8}+\frac{\sqrt3}{2}a=0\ \Longrightarrow\ -66+9a+4\sqrt3\,a=0.$$ Step 3 - Solve. $a=\dfrac{66}{9+4\sqrt3}$, and $A''=\frac98+\frac{\sqrt3}{2}>0$, so it is a minimum. Answer: (B)