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Question Number 119240 by benjo_mathlover last updated on 23/Oct/20
find max and min value of   f(x,y) = 4x^2 +8xy+9y^2 −8x−24y+4
$${find}\:{max}\:{and}\:{min}\:{value}\:{of}\: \\ $$$${f}\left({x},{y}\right)\:=\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{xy}+\mathrm{9}{y}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{24}{y}+\mathrm{4}\: \\ $$
Answered by 1549442205PVT last updated on 23/Oct/20
f(x,y) = 4x^2 +8xy+9y^2 −8x−24y+4   =[4x^2 +2.2x(2y−2)+4y^2 −8y+4]+5y^2 −16y  =(2x+2y−2)^2 +5y^2 −16y  =4(x+y−1)^2 +5(y−(8/5))^2 −((64)/5)≥((−64)/5)  ⇒f_(min) (x,y)=((−64)/5) when   { ((y−8/5=0)),((x+y−1=0)) :}⇔ { ((x=−3/5)),((y=8/5)) :}  f_(max) (x,y)=+∞
$${f}\left({x},{y}\right)\:=\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{xy}+\mathrm{9}{y}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{24}{y}+\mathrm{4}\: \\ $$$$=\left[\mathrm{4x}^{\mathrm{2}} +\mathrm{2}.\mathrm{2x}\left(\mathrm{2y}−\mathrm{2}\right)+\mathrm{4y}^{\mathrm{2}} −\mathrm{8y}+\mathrm{4}\right]+\mathrm{5y}^{\mathrm{2}} −\mathrm{16y} \\ $$$$=\left(\mathrm{2x}+\mathrm{2y}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{5y}^{\mathrm{2}} −\mathrm{16y} \\ $$$$=\mathrm{4}\left(\mathrm{x}+\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{5}\left(\mathrm{y}−\frac{\mathrm{8}}{\mathrm{5}}\right)^{\mathrm{2}} −\frac{\mathrm{64}}{\mathrm{5}}\geqslant\frac{−\mathrm{64}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{f}_{\mathrm{min}} \left(\mathrm{x},\mathrm{y}\right)=\frac{−\mathrm{64}}{\mathrm{5}}\:\mathrm{when} \\ $$$$\begin{cases}{\mathrm{y}−\mathrm{8}/\mathrm{5}=\mathrm{0}}\\{\mathrm{x}+\mathrm{y}−\mathrm{1}=\mathrm{0}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{x}=−\mathrm{3}/\mathrm{5}}\\{\mathrm{y}=\mathrm{8}/\mathrm{5}}\end{cases} \\ $$$$\mathrm{f}_{\mathrm{max}} \left(\mathrm{x},\mathrm{y}\right)=+\infty \\ $$
Answered by ebi last updated on 23/Oct/20
  f_x =(d/dx)f  f_x =8x+8y−8  f_x =0  8x+8y−8=0...... (1)    f_y =(d/dy)f  f_y =8x+18y−24  f_y =0  8x+18y−24=0......(2)    (1) −(2): −10y=−16→y=(8/5)  ∴ 8x+8((8/5))=8→x=−(3/5)  (−(3/5),(8/5)) when f_x =0 and f_y =0    D−Test  D=f_(xx) ∙f_(yy) −(f_(xy) )^2   f_(xx) =8  f_(yy) =18  f_(xy) =8  D=8(18)−8^2 =80  since D=80>0 and f_(xx) =8>0,  then f has a minimum at (−(3/5),(8/5))
$$ \\ $$$${f}_{{x}} =\frac{{d}}{{dx}}{f} \\ $$$${f}_{{x}} =\mathrm{8}{x}+\mathrm{8}{y}−\mathrm{8} \\ $$$${f}_{{x}} =\mathrm{0} \\ $$$$\mathrm{8}{x}+\mathrm{8}{y}−\mathrm{8}=\mathrm{0}……\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$${f}_{{y}} =\frac{{d}}{{dy}}{f} \\ $$$${f}_{{y}} =\mathrm{8}{x}+\mathrm{18}{y}−\mathrm{24} \\ $$$${f}_{{y}} =\mathrm{0} \\ $$$$\mathrm{8}{x}+\mathrm{18}{y}−\mathrm{24}=\mathrm{0}……\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:−\left(\mathrm{2}\right):\:−\mathrm{10}{y}=−\mathrm{16}\rightarrow{y}=\frac{\mathrm{8}}{\mathrm{5}} \\ $$$$\therefore\:\mathrm{8}{x}+\mathrm{8}\left(\frac{\mathrm{8}}{\mathrm{5}}\right)=\mathrm{8}\rightarrow{x}=−\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\left(−\frac{\mathrm{3}}{\mathrm{5}},\frac{\mathrm{8}}{\mathrm{5}}\right)\:{when}\:{f}_{{x}} =\mathrm{0}\:{and}\:{f}_{{y}} =\mathrm{0} \\ $$$$ \\ $$$${D}−{Test} \\ $$$${D}={f}_{{xx}} \centerdot{f}_{{yy}} −\left({f}_{{xy}} \right)^{\mathrm{2}} \\ $$$${f}_{{xx}} =\mathrm{8} \\ $$$${f}_{{yy}} =\mathrm{18} \\ $$$${f}_{{xy}} =\mathrm{8} \\ $$$${D}=\mathrm{8}\left(\mathrm{18}\right)−\mathrm{8}^{\mathrm{2}} =\mathrm{80} \\ $$$${since}\:{D}=\mathrm{80}>\mathrm{0}\:{and}\:{f}_{{xx}} =\mathrm{8}>\mathrm{0}, \\ $$$${then}\:{f}\:{has}\:{a}\:{minimum}\:{at}\:\left(−\frac{\mathrm{3}}{\mathrm{5}},\frac{\mathrm{8}}{\mathrm{5}}\right) \\ $$

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