JEE Main 2022 (Online) 30th July Evening ShiftQuestion 1
An air duct is made of a thick metallic sheet. The duct is open from front and back. Its cross-section is a trapezium of base angles $60^\circ$ and area $1.5$ m$^2$. Its length is $1$ m. [figure not transcribed: trapezium cross-section with height $x$ and base angles $60^\circ$] If its height $x=x_0$ (in metres) so that the area of the metallic sheet used for its construction is minimum, then $16(x_0)^4$ is equal to ________.
Differentiation and its Application
Answer$27$
Solution
Concept. The duct has length $1$, so the sheet area equals the perimeter of the cross-section. Express it in $x$ using the fixed area. Formulas used. (1) With base angle $60^\circ$ and height $x$: each slant side is $\frac{2x}{\sqrt3}$ and sticks out $\frac{x}{\sqrt3}$ horizontally(2) Area of trapezium $=\frac12(\text{sum of parallel sides})\times\text{height}$Step 1 - Cross-section.Step 2 - Sides. Let $b$ be the longer base. By (1), the top is $b-\frac{2x}{\sqrt3}$. By (2), $$x\left(b-\frac{x}{\sqrt3}\right)=\frac32\ \Longrightarrow\ b=\frac{3}{2x}+\frac{x}{\sqrt3}.$$ Step 3 - Perimeter. $$S=b+\left(b-\frac{2x}{\sqrt3}\right)+\frac{4x}{\sqrt3}=2b+\frac{2x}{\sqrt3}=\frac3x+\frac{4x}{\sqrt3}.$$ Step 4 - Minimise. $$S'=-\frac{3}{x^2}+\frac{4}{\sqrt3}=0\ \Longrightarrow\ x^2=\frac{3\sqrt3}{4}.$$ $S''=\frac{6}{x^3}>0$, so it is a minimum. Step 5 - Value. $x_0^4=\frac{27}{16}$, so $16x_0^4=27$. Answer: 27