Question #314807

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.


1
Expert's answer
2022-03-21T00:20:11-0400

T(x,y)=x2+xy+y26x+2{Tx=0Ty=0{2x+y6=0x+2y=0{x=4y=2T(4,2)=42+4(2)+(2)264+2=10T(x,3)=x23x+96x+2=x29x+11T(0,3)=11,T(5,3)=9,T(4.5,3)=9.25T(x,3)=x2+3x+96x+2=x23x+11T(0,3)=11,T(5,3)=21,T(1.5,3)=8.75T(0,y)=y2+2T(0,3)=11,T(0,3)=11,T(0,0)=2T(5,y)=25+5y+y230+2=y2+5y3T(5,3)=9,T(5,3)=21,T(5,2.5)=9.25T\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

minT=10,maxT=21\min T=-10,\max T=21


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