Answer to Question #314807 in Calculus for mimi

Question

Find the absolute maxima and minima of the function on the given domain

T(x, y) = x

2 + xy + y

2 − 6x + 2 on the rectangular plane 0 ≤ x ≤ 5, −3 ≤ y ≤ 3.

Answer 1

$T\left( x,y \right) =x^2+xy+y^2-6x+2\\\left\{ \begin{array}{c} T'_x=0\\ T'_y=0\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} 2x+y-6=0\\ x+2y=0\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} x=4\\ y=-2\\\end{array} \right. \\T\left( 4,-2 \right) =4^2+4\cdot \left( -2 \right) +\left( -2 \right) ^2-6\cdot 4+2=-10\\T\left( x,-3 \right) =x^2-3x+9-6x+2=x^2-9x+11\\T\left( 0,-3 \right) =11,T\left( 5,-3 \right) =-9,T\left( 4.5,-3 \right) =-9.25\\T\left( x,3 \right) =x^2+3x+9-6x+2=x^2-3x+11\\T\left( 0,3 \right) =11,T\left( 5,3 \right) =21,T\left( 1.5,3 \right) =8.75\\T\left( 0,y \right) =y^2+2\\T\left( 0,-3 \right) =11,T\left( 0,3 \right) =11,T\left( 0,0 \right) =2\\T\left( 5,y \right) =25+5y+y^2-30+2=y^2+5y-3\\T\left( 5,-3 \right) =-9,T\left( 5,3 \right) =21,T\left( 5,-2.5 \right) =-9.25$

Thus

$\min T=-10,\max T=21$