Question Number 119240 by benjo_mathlover last updated on 23/Oct/20
$${find}\:{max}\:{and}\:{min}\:{value}\:{of}\: \\ $$$${f}\left({x},{y}\right)\:=\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{xy}+\mathrm{9}{y}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{24}{y}+\mathrm{4}\: \\ $$
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)=+∞](https://www.tinkutara.com/question/Q119242.png)
$${f}\left({x},{y}\right)\:=\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{xy}+\mathrm{9}{y}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{24}{y}+\mathrm{4}\: \\ $$$$=\left[\mathrm{4x}^{\mathrm{2}} +\mathrm{2}.\mathrm{2x}\left(\mathrm{2y}−\mathrm{2}\right)+\mathrm{4y}^{\mathrm{2}} −\mathrm{8y}+\mathrm{4}\right]+\mathrm{5y}^{\mathrm{2}} −\mathrm{16y} \\ $$$$=\left(\mathrm{2x}+\mathrm{2y}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{5y}^{\mathrm{2}} −\mathrm{16y} \\ $$$$=\mathrm{4}\left(\mathrm{x}+\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{5}\left(\mathrm{y}−\frac{\mathrm{8}}{\mathrm{5}}\right)^{\mathrm{2}} −\frac{\mathrm{64}}{\mathrm{5}}\geqslant\frac{−\mathrm{64}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{f}_{\mathrm{min}} \left(\mathrm{x},\mathrm{y}\right)=\frac{−\mathrm{64}}{\mathrm{5}}\:\mathrm{when} \\ $$$$\begin{cases}{\mathrm{y}−\mathrm{8}/\mathrm{5}=\mathrm{0}}\\{\mathrm{x}+\mathrm{y}−\mathrm{1}=\mathrm{0}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{x}=−\mathrm{3}/\mathrm{5}}\\{\mathrm{y}=\mathrm{8}/\mathrm{5}}\end{cases} \\ $$$$\mathrm{f}_{\mathrm{max}} \left(\mathrm{x},\mathrm{y}\right)=+\infty \\ $$
Answered by ebi last updated on 23/Oct/20
$$ \\ $$$${f}_{{x}} =\frac{{d}}{{dx}}{f} \\ $$$${f}_{{x}} =\mathrm{8}{x}+\mathrm{8}{y}−\mathrm{8} \\ $$$${f}_{{x}} =\mathrm{0} \\ $$$$\mathrm{8}{x}+\mathrm{8}{y}−\mathrm{8}=\mathrm{0}……\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$${f}_{{y}} =\frac{{d}}{{dy}}{f} \\ $$$${f}_{{y}} =\mathrm{8}{x}+\mathrm{18}{y}−\mathrm{24} \\ $$$${f}_{{y}} =\mathrm{0} \\ $$$$\mathrm{8}{x}+\mathrm{18}{y}−\mathrm{24}=\mathrm{0}……\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:−\left(\mathrm{2}\right):\:−\mathrm{10}{y}=−\mathrm{16}\rightarrow{y}=\frac{\mathrm{8}}{\mathrm{5}} \\ $$$$\therefore\:\mathrm{8}{x}+\mathrm{8}\left(\frac{\mathrm{8}}{\mathrm{5}}\right)=\mathrm{8}\rightarrow{x}=−\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\left(−\frac{\mathrm{3}}{\mathrm{5}},\frac{\mathrm{8}}{\mathrm{5}}\right)\:{when}\:{f}_{{x}} =\mathrm{0}\:{and}\:{f}_{{y}} =\mathrm{0} \\ $$$$ \\ $$$${D}−{Test} \\ $$$${D}={f}_{{xx}} \centerdot{f}_{{yy}} −\left({f}_{{xy}} \right)^{\mathrm{2}} \\ $$$${f}_{{xx}} =\mathrm{8} \\ $$$${f}_{{yy}} =\mathrm{18} \\ $$$${f}_{{xy}} =\mathrm{8} \\ $$$${D}=\mathrm{8}\left(\mathrm{18}\right)−\mathrm{8}^{\mathrm{2}} =\mathrm{80} \\ $$$${since}\:{D}=\mathrm{80}>\mathrm{0}\:{and}\:{f}_{{xx}} =\mathrm{8}>\mathrm{0}, \\ $$$${then}\:{f}\:{has}\:{a}\:{minimum}\:{at}\:\left(−\frac{\mathrm{3}}{\mathrm{5}},\frac{\mathrm{8}}{\mathrm{5}}\right) \\ $$