6. Application of derivatives — ગણિત ધોરણ 12 સાયન્સ — Question

$ \Rightarrow \,\frac{{{\text{dy}}}}{{{\text{dx}}}}\,\, = \,\,\frac{1}{{40}}\,(12{x^3} + 24{x^2} - 36x)$

અને $\frac{{{{\text{d}}^{\text{2}}}y}}{{d{x^2}}}\,\, = \,\,\frac{1}{{40}}\,(36{x^2} + 48x - 36)$

હવે $\frac{{{\text{dy}}}}{{{\text{dx}}}}\,\, = \,\,0\,\, \Rightarrow \,{x^3} + 2{x^2} - 3x = 0$

અથવા ${\text{x(x - 1)}}\,{\text{(x}}\, + \,{\text{3)}}\,\, = \,{\text{0}}$ અથવા ${\text{x}}\, = \,\,{\text{0,}}\,{\text{1}}\,\,{\text{ - 3}}$

${\text{x}}\, = \,{\text{0}}\,\,$ આગળ $\frac{{{{\text{d}}^{\text{2}}}y}}{{d{x^2}}}\,\, = \,\,{\text{ - 36}}\,\, < \,\,{\text{0}}\,\,\,\,\therefore \,\,{\text{x}}\, = \,{\text{0}} $ આગળ ${\text{y}}$ મહતમ છે. 

$ \Rightarrow $ ઉ.દા. આપેલ વિધેય $\frac{{\text{1}}}{{\text{y}}},\,\,{\text{x}}\, = \,{\text{0}}$ આગળ ન્યૂનતમ છે. 

વિધેયનું ન્યૂનતમ મૂલ્ય $ = \,\,\frac{{40}}{{60}}\,\, = \,\,\frac{2}{3}$