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Maximum and minimum values

Find the maximum and minimum values of x 3 + 3xy 2 – 3x 2 – 3y 2 + 7.

1 Answer
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Let, $f(x,y)=x^3+3xy^2-3x^2-3y^2+7 \\ \; \\$

The stationary points are given by $\dfrac{\partial u}{\partial x}=0$ and $\dfrac{\partial u}{\partial y}=0$

$\\ \; \\ \therefore \dfrac{\partial u}{\partial x} \;=\; 3x^2+3y^2-6x=0 \; \; \; \therefore x^2+y^2-2x=0 \; \; \ldots(i) \\$

Also, $\\ \; \\ \therefore \dfrac{\partial u}{\partial y} \;=\; =6xy-6y=0 \; \; \; \therefore y(x-1)=0 \;\; \;\ldots (ii) \\$

From equation (ii),y=0 and x=1

$\\$

Substituting y=0 in equation (i),

$\\ \therefore x^2-2x=0 \; \; \; \therefore x(x-2)=0 \; \; \; \therefore x=0 \; or \; x=2 \\ \; \\ \; \\$

$\therefore$Stationary points are (0,0),(2,0)

$\\$

Substituting x=1 in equation (i),

$\\ \therefore 1+y^2-2=0 ∴y^2=1 ∴y=-1 \; or \; y=1 \\$ $\therefore$ stationary points are (1,-1),(1,1). $\$ Now, $r\;=\; \dfrac{\partial^2 f}{\partial x^2}=6x-6, \;\; s= \dfrac{\partial^2 f}{\partial x \partial y }=6y \; \; and \; \; t= \dfrac{\partial^2 f}{\partial y^2} \;=\; 6x-6 \ \; \ \; \$ $\ (i) \; \; At \; \; (0,0), \; rt-s^2 \;= \; (6x-6)(6x-6)-36y^2 \;=\; (-6)(-6)-0 \;= \;36>0 \; and \; r=-6 \;<\;0$ Hence,at (0,0),function is maximum and $f_{max}=7$ $\$ $(ii) \; At \; (2,0), \; rt-s^2 \;=\;(6x-6)(6x-6)-36y^2 \;= \; (6)(6)-0 \;=\;36>0 \; and \; r=6 \;> \;0 \ \; \$ Hence,at (2,0),function is minimum and $f_{min}=3$ $\ \; \ \; \ (iii) \; At \; (1,1) \; and \; (1,-1), \; rt-s^2 \;=\;(6x-6)(6x-6)-36y^2 \; =(0)(0)-(±6)^2 \; =-36 \; <0 \$

Hence,at (1,1)and (1,-1),function is neither maximum nor minimum.

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