Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1

Marks : 6 M

Year : May 2013

Let $ u \;=\; x^3 + y^3 -3axy \\ $

Diff partially w.r.t. x,

$ \therefore \dfrac{\partial u}{\partial x} \;=\; 3x^2 - 3ay \\ $

Equating it to zero,

$ \therefore 3x^2 - 3ay \;= \; 0 \\ x^2 \;=\; ay \; \; \; \ldots (i) \\ \; \\ $

Diff. partially w.r.t. y,

$ \therefore \dfrac{\partial u}{\partial y} \;=\; 3y^2 - 3ax \\ $

Equating it to zero,

$ \therefore 3y^2 - 3ax \;= \; 0 \\ y^2 \;=\; ax \; \; \; \ldots (ii) \\ \; \\ $

On squaring equation (i), $ \therefore x^4 \;=\; a^2 y^2 \\ \; \\ \therefore x^4 \;=\; a^2 (ax) \; \; \ldots From \; (ii) \\ \; \\ \therefore x^4 \;=\; a^3x \\ \; \\ \therefore x^3 \;=\; a^3 \\ \; \\ \therefore x=a \\ \; \\ \; \\ $

Substituting in equation (i)

$ \therefore a^2 \;=\; ay \; \; \; \; \therefore y=a \\ \; \\ \; \\ $

$ \left[ \begin{array}{c} \dfrac{\partial^2 u}{\partial x^2} \;=\; 6x \;=\; 6a \;\gt\; 0 \; \; \; \dfrac{\partial^2 u}{\partial x^2} \;=\; 6y \;=\; 6a \;\gt\; 0 \; \; \; \; \; \; \\ \because \dfrac{\partial^2 u}{\partial x^2} \; \; \& \; \; \dfrac{\partial^2 u}{\partial y^2} \; \gt \; 0 \; \; \; Hence \; it's \; a \; minima \end{array}\right] $