Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 83871 by jagoll last updated on 07/Mar/20

If equation { (((√(x^2 +y^2 ))+(√((x−4)^2 +y^2 ))+(√(x^2 +(y−3)^2 ))+(√((x−4)^2 +(y−3)^2 ))=10)),((x+2y= 5z)) :} has solution is (a,b,c). find a+2b+3c

$$\mathrm{If}\:\mathrm{equation}\: \\ $$$$\begin{cases}{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }+\sqrt{\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{10}}\\{\mathrm{x}+\mathrm{2y}=\:\mathrm{5z}}\end{cases} \\ $$$$\mathrm{has}\:\mathrm{solution}\:\mathrm{is}\:\left(\mathrm{a},\mathrm{b},\mathrm{c}\right).\: \\ $$$$\mathrm{find}\:\mathrm{a}+\mathrm{2b}+\mathrm{3c}\: \\ $$

Commented by mr W last updated on 07/Mar/20

way 1: let f(x,y)=(√(x^2 +y^2 ))+(√((x−4)^2 +y^2 ))+(√(x^2 +(y−3)^2 ))+(√((x−4)^2 +(y−3)^2 )) with (∂f/∂x)=0, (∂f/∂y)=0 you can get f(x,y)_(min) =10 at x=2, y=(3/2) ⇒f(x,y)=10 has only one solution: x=2 and y=(3/2). way 2: LHS=(√(x^2 +y^2 ))+(√((x−4)^2 +y^2 ))+(√(x^2 +(y−3)^2 ))+(√((x−4)^2 +(y−3)^2 )) is the sum of distances from a point (x,y) to the 4 vertices of a rectangle with side lengthes 4 and 3. the smallest sum is when this point is at the center of the rectangle which lies at (2,(3/2)). the smallest sum is the lengthes of both diagonals, i.e. 2×5=10. that means LHS≥10. for LHS=10 there is only one solution: x=2, y=(3/2).

$${way}\:\mathrm{1}: \\ $$$${let}\:{f}\left({x},{y}\right)=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }+\sqrt{\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${with}\:\frac{\partial{f}}{\partial{x}}=\mathrm{0},\:\frac{\partial{f}}{\partial{y}}=\mathrm{0} \\ $$$${you}\:{can}\:{get}\:{f}\left({x},{y}\right)_{{min}} =\mathrm{10}\:{at}\:{x}=\mathrm{2},\:{y}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x},{y}\right)=\mathrm{10}\:{has}\:{only}\:{one}\:{solution}: \\ $$$${x}=\mathrm{2}\:{and}\:{y}=\frac{\mathrm{3}}{\mathrm{2}}. \\ $$$$ \\ $$$${way}\:\mathrm{2}: \\ $$$${LHS}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }+\sqrt{\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${is}\:{the}\:{sum}\:{of}\:{distances}\:{from}\:{a}\:{point} \\ $$$$\left({x},{y}\right)\:{to}\:{the}\:\mathrm{4}\:{vertices}\:{of}\:{a}\:{rectangle} \\ $$$${with}\:{side}\:{lengthes}\:\mathrm{4}\:{and}\:\mathrm{3}. \\ $$$${the}\:{smallest}\:{sum}\:{is}\:{when}\:{this}\:{point} \\ $$$${is}\:{at}\:{the}\:{center}\:{of}\:{the}\:{rectangle}\:{which} \\ $$$${lies}\:{at}\:\left(\mathrm{2},\frac{\mathrm{3}}{\mathrm{2}}\right).\:{the}\:{smallest}\:{sum}\:{is}\:{the} \\ $$$${lengthes}\:{of}\:{both}\:{diagonals},\:{i}.{e}.\:\mathrm{2}×\mathrm{5}=\mathrm{10}. \\ $$$${that}\:{means}\:{LHS}\geqslant\mathrm{10}.\:{for}\:{LHS}=\mathrm{10} \\ $$$${there}\:{is}\:{only}\:{one}\:{solution}: \\ $$$${x}=\mathrm{2},\:{y}=\frac{\mathrm{3}}{\mathrm{2}}. \\ $$

Commented by mr W last updated on 07/Mar/20

x=a=2, y=b=(3/2), z=c=1 a+2b+3z=2+3+3=8

$${x}={a}=\mathrm{2},\:{y}={b}=\frac{\mathrm{3}}{\mathrm{2}},\:{z}={c}=\mathrm{1} \\ $$$${a}+\mathrm{2}{b}+\mathrm{3}{z}=\mathrm{2}+\mathrm{3}+\mathrm{3}=\mathrm{8} \\ $$

Commented by jagoll last updated on 07/Mar/20

how to got it sir?

$$\mathrm{how}\:\mathrm{to}\:\mathrm{got}\:\mathrm{it}\:\mathrm{sir}? \\ $$

Commented by jagoll last updated on 07/Mar/20

why your answer blank sir

$$\mathrm{why}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{blank}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com