Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 157247 by john_santu last updated on 21/Oct/21

F(x,y)=x^2 −2xy+6y^2 −12x+2y+45 find x &y such that F(x,y) minimum

$${F}\left({x},{y}\right)={x}^{\mathrm{2}} −\mathrm{2}{xy}+\mathrm{6}{y}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{2}{y}+\mathrm{45} \\ $$ $${find}\:{x}\:\&{y}\:{such}\:{that}\:{F}\left({x},{y}\right)\:{minimum} \\ $$

Answered by FongXD last updated on 21/Oct/21

Given: f(x,y)=x^2 −2xy+6y^2 −12x+2y+45 ⇔ f(x,y)=(x^2 +y^2 +36−2xy−12x+12y)+(5y^2 −10y+9) ⇔ f(x,y)=(x−y−6)^2 +5(y−1)^2 +4 since (x−y−6)^2 +5(y−1)^2 ≥0, ∀x,y∈R therefore, Min[f(x,y)]=4, when y=1 and x=7 so. determinant (((Min[f(x,y)]=4 which occurs when x=7 and y=1)))

$$\mathrm{Given}:\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{2xy}+\mathrm{6y}^{\mathrm{2}} −\mathrm{12x}+\mathrm{2y}+\mathrm{45} \\ $$ $$\Leftrightarrow\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{36}−\mathrm{2xy}−\mathrm{12x}+\mathrm{12y}\right)+\left(\mathrm{5y}^{\mathrm{2}} −\mathrm{10y}+\mathrm{9}\right) \\ $$ $$\Leftrightarrow\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{x}−\mathrm{y}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{5}\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4} \\ $$ $$\mathrm{since}\:\left(\mathrm{x}−\mathrm{y}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{5}\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{0},\:\forall\mathrm{x},\mathrm{y}\in\mathbb{R} \\ $$ $$\mathrm{therefore},\:\mathrm{Min}\left[\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\right]=\mathrm{4},\:\mathrm{when}\:\mathrm{y}=\mathrm{1}\:\mathrm{and}\:\mathrm{x}=\mathrm{7} \\ $$ $$\mathrm{so}.\:\begin{array}{|c|}{\mathrm{Min}\left[\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\right]=\mathrm{4}\:\mathrm{which}\:\mathrm{occurs}\:\mathrm{when}\:\mathrm{x}=\mathrm{7}\:\mathrm{and}\:\mathrm{y}=\mathrm{1}}\\\hline\end{array} \\ $$

Answered by mr W last updated on 21/Oct/21

x^2 −2xy+6y^2 −12x+2y+45=k x^2 −2(y+6)x+6y^2 +2y+45−k=0 (y+6)^2 −(6y^2 +2y+45−k)≥0 5y^2 −10y+9−k≤0 10^2 −4×5(9−k)≥0 k≥4 ⇒F(x,y)_(min) =4 y^2 −2y+1=0 ⇒y=1 x^2 −14x+49=0 ⇒x=7

$${x}^{\mathrm{2}} −\mathrm{2}{xy}+\mathrm{6}{y}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{2}{y}+\mathrm{45}={k} \\ $$ $${x}^{\mathrm{2}} −\mathrm{2}\left({y}+\mathrm{6}\right){x}+\mathrm{6}{y}^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{45}−{k}=\mathrm{0} \\ $$ $$\left({y}+\mathrm{6}\right)^{\mathrm{2}} −\left(\mathrm{6}{y}^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{45}−{k}\right)\geqslant\mathrm{0} \\ $$ $$\mathrm{5}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{9}−{k}\leqslant\mathrm{0} \\ $$ $$\mathrm{10}^{\mathrm{2}} −\mathrm{4}×\mathrm{5}\left(\mathrm{9}−{k}\right)\geqslant\mathrm{0} \\ $$ $${k}\geqslant\mathrm{4} \\ $$ $$\Rightarrow{F}\left({x},{y}\right)_{{min}} =\mathrm{4} \\ $$ $${y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}=\mathrm{0}\:\Rightarrow{y}=\mathrm{1} \\ $$ $${x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{49}=\mathrm{0}\:\Rightarrow{x}=\mathrm{7} \\ $$

Answered by qaz last updated on 21/Oct/21

{ (((∂F/∂x)=2x−2y−12=0)),(((∂F/∂y)=−2x+12y+2=0)) :} ⇒M=(x,y)=(7,1) A=(∂^2 F/∂x^2 )∣_M =2 B=(∂^2 F/∂y^2 )∣_M =12 C=(∂^2 F/(∂x∂y))∣_M =−2 ∵ A>0 AB−C^2 >0 ∴ MinF(x,y)=F(7,1)=4

$$\begin{cases}{\frac{\partial\mathrm{F}}{\partial\mathrm{x}}=\mathrm{2x}−\mathrm{2y}−\mathrm{12}=\mathrm{0}}\\{\frac{\partial\mathrm{F}}{\partial\mathrm{y}}=−\mathrm{2x}+\mathrm{12y}+\mathrm{2}=\mathrm{0}}\end{cases}\:\Rightarrow\mathrm{M}=\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{7},\mathrm{1}\right) \\ $$ $$\mathrm{A}=\frac{\partial^{\mathrm{2}} \mathrm{F}}{\partial\mathrm{x}^{\mathrm{2}} }\mid_{\mathrm{M}} =\mathrm{2}\:\:\:\mathrm{B}=\frac{\partial^{\mathrm{2}} \mathrm{F}}{\partial\mathrm{y}^{\mathrm{2}} }\mid_{\mathrm{M}} =\mathrm{12}\:\:\:\:\mathrm{C}=\frac{\partial^{\mathrm{2}} \mathrm{F}}{\partial\mathrm{x}\partial\mathrm{y}}\mid_{\mathrm{M}} =−\mathrm{2} \\ $$ $$\because\:\:\:\mathrm{A}>\mathrm{0}\:\:\:\:\:\mathrm{AB}−\mathrm{C}^{\mathrm{2}} >\mathrm{0} \\ $$ $$\therefore\:\:\mathrm{MinF}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{F}\left(\mathrm{7},\mathrm{1}\right)=\mathrm{4} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com