Q.$f(x) = (x-1)(x-2)(x-3)$, $x \in [0,4]$, find 'c' if LMVT can be applied. OR A rod of 108 meters long is bent to form a rectangle. Find its dimensions if the area is maximum.

Part 1: apply the Lagrange Mean Value Theorem formula $f'(c)=\dfrac{f(b)-f(a)}{b-a}$. Part 2 (OR): maximize area $=x(54-x)$.

Part 1 — LMVT for $f(x)=(x-1)(x-2)(x-3)$, $x\in[0,4]$:

$f$ is a polynomial, hence continuous on $[0,4]$ and differentiable on $(0,4)$ — LMVT applies.

Expanding: $f(x) = x^3-6x^2+11x-6$

$f(4) = 64-96+44-6 = 6$, $\quad f(0) = -6$

$\dfrac{f(4)-f(0)}{4-0} = \dfrac{6-(-6)}{4} = 3$

$f'(x) = 3x^2-12x+11$. Set $f'(c)=3$:

$3c^2-12c+11=3 \implies 3c^2-12c+8=0$

$c = \dfrac{12\pm\sqrt{144-96}}{6} = \dfrac{12\pm\sqrt{48}}{6} = \dfrac{12\pm4\sqrt3}{6} = 2\pm\dfrac{2\sqrt3}{3}$

Numerically, $c\approx3.155$ or $c\approx0.845$ — both lie in $(0,4)$, so both are valid values of $c$.

Part 2 (OR) — Rod of 108 m bent into a rectangle, maximum area:

Let the sides be $x$ and $y$. Perimeter: $2x+2y=108 \implies y=54-x$.

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