JEE Main 2021 (Online) 27th August Evening ShiftQuestion 1
A box open from top is made from a rectangular sheet of dimension $a\times b$ by cutting squares each of side $x$ from each of the four corners and folding up the flaps. If the volume of the box is maximum, then $x$ is equal to :
Differentiation and its Application
Solution
Concept. $V=x(a-2x)(b-2x)$; $V'$ is a quadratic with two roots, and only the smaller one lies in the allowed range $0<x<\frac a2$ (take $a\le b$). Formulas used. (1) $V'(x)=12x^2-4(a+b)x+ab$Step 1 - Roots. By (1), $$x=\frac{4(a+b)\pm\sqrt{16(a+b)^2-48ab}}{24}=\frac{a+b\pm\sqrt{a^2+b^2-ab}}{6}.$$ Step 2 - Which root. $V'(0)=ab>0$ and $V'\left(\frac a2\right)=3a^2-2a(a+b)+ab=a(a-b)\le0$. So exactly one root lies in $\left(0,\frac a2\right)$, the smaller one, and $V'$ goes $+\to-$ there: the maximum. Step 3 - Answer. $x=\dfrac{a+b-\sqrt{a^2+b^2-ab}}{6}$. Answer: (A)