Menu Close

find-minimum-value-of-function-y-x-6-2-25-x-6-2-121-

Tinku Tara 0 Comments
Pin




Question Number 114922 by bemath last updated on 22/Sep/20
find minimum value of function y=(√((x+6)^2 +25)) +(√((x−6)^2 +121))
$${find}\:{minimum}\:{value}\:{of}\:{function} \\ $$$${y}=\sqrt{\left({x}+\mathrm{6}\right)^{\mathrm{2}} +\mathrm{25}}\:+\sqrt{\left({x}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{121}} \\ $$
Answered by john santu last updated on 22/Sep/20
you want to find the point on the x−axis such that the sum of its distance from the points (−6,5) and (6,11) is minimal. consider the symetric point of (−6,5) with respect to the axis, that is (−6,−5). the line through (−6,−5) and (6,11) has equation 4x−3y+9=0 and it intersects the x−axis at x=−(9/4). The minimum value is therefore y=(√((−(9/4)+6)^2 +25))+(√((−(9/6)−6)^2 +121)) y = ((25)/4) + ((55)/4) = 20 ∴
$${you}\:{want}\:{to}\:{find}\:{the}\:{point}\:{on}\:{the}\: \\ $$$${x}−{axis}\:{such}\:{that}\:{the}\:{sum}\:{of}\:{its}\: \\ $$$${distance}\:{from}\:{the}\:{points}\:\left(−\mathrm{6},\mathrm{5}\right)\:{and} \\ $$$$\left(\mathrm{6},\mathrm{11}\right)\:{is}\:{minimal}. \\ $$$${consider}\:{the}\:{symetric}\:{point}\:{of}\:\left(−\mathrm{6},\mathrm{5}\right) \\ $$$${with}\:{respect}\:{to}\:{the}\:{axis},\:{that}\:{is}\:\left(−\mathrm{6},−\mathrm{5}\right). \\ $$$${the}\:{line}\:{through}\:\left(−\mathrm{6},−\mathrm{5}\right)\:{and}\:\left(\mathrm{6},\mathrm{11}\right) \\ $$$${has}\:{equation}\:\mathrm{4}{x}−\mathrm{3}{y}+\mathrm{9}=\mathrm{0} \\ $$$${and}\:{it}\:{intersects}\:{the}\:{x}−{axis}\:{at}\: \\ $$$${x}=−\frac{\mathrm{9}}{\mathrm{4}}.\:{The}\:{minimum}\:{value}\: \\ $$$${is}\:{therefore}\:{y}=\sqrt{\left(−\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{6}\right)^{\mathrm{2}} +\mathrm{25}}+\sqrt{\left(−\frac{\mathrm{9}}{\mathrm{6}}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{121}} \\ $$$${y}\:=\:\frac{\mathrm{25}}{\mathrm{4}}\:+\:\frac{\mathrm{55}}{\mathrm{4}}\:=\:\mathrm{20}\:\therefore \\ $$
Commented by bemath last updated on 22/Sep/20
gave kudos
$${gave}\:{kudos} \\ $$
Answered by bobhans last updated on 22/Sep/20
with calculus y = (√((x+6)^2 +25)) + (√((x−6)^2 +121)) y ′=(((x+6))/( (√((x+6)^2 +25)))) + (((x−6))/( (√((x−6)^2 +121)))) = 0 (x+6)(√((x−6)^2 +121)) +(x−6)(√((x+6)^2 +25)) = 0 (x+6)(√((x−6)^2 +121)) = (6−x)(√((x+6)^2 +25)) ((√((x−6)^2 +121))/( (√((x+6)^2 +25)))) = ((6−x)/(x+6)) (((x−6)^2 +121)/((x+6)^2 +25)) = (((x−6)^2 )/((x+6)^2 )) (x^2 −36)^2 +121(x+6)^2 =(x^2 −36)^2 +25(x−6)^2 (11x+66)^2 =(5x−30)^2 (16x+36)(6x+96)=0 { ((x=−(9/4))),((x=−16)) :} for x=−(9/4) y=(√((−(9/4)+6)^2 +25)) +(√((−(9/4)−6)^2 +121)) y=(√((625)/(16))) + (√((3025)/(16))) = ((25)/4)+((55)/4)=20 for x=−16 y=(√((−16+6)^2 +25)) +(√((−16−6)^2 +121)) y=(√(125)) +(√(605)) = 35.78 therefore minimum value is 20
$${with}\:{calculus} \\ $$$${y}\:=\:\sqrt{\left({x}+\mathrm{6}\right)^{\mathrm{2}} +\mathrm{25}}\:+\:\sqrt{\left({x}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{121}} \\ $$$${y}\:'=\frac{\left({x}+\mathrm{6}\right)}{\:\sqrt{\left({x}+\mathrm{6}\right)^{\mathrm{2}} +\mathrm{25}}}\:+\:\frac{\left({x}−\mathrm{6}\right)}{\:\sqrt{\left({x}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{121}}}\:=\:\mathrm{0} \\ $$$$\left({x}+\mathrm{6}\right)\sqrt{\left({x}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{121}}\:+\left({x}−\mathrm{6}\right)\sqrt{\left({x}+\mathrm{6}\right)^{\mathrm{2}} +\mathrm{25}}\:=\:\mathrm{0} \\ $$$$\left({x}+\mathrm{6}\right)\sqrt{\left({x}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{121}}\:=\:\left(\mathrm{6}−{x}\right)\sqrt{\left({x}+\mathrm{6}\right)^{\mathrm{2}} +\mathrm{25}} \\ $$$$\frac{\sqrt{\left({x}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{121}}}{\:\sqrt{\left({x}+\mathrm{6}\right)^{\mathrm{2}} +\mathrm{25}}}\:=\:\frac{\mathrm{6}−{x}}{{x}+\mathrm{6}} \\ $$$$\frac{\left({x}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{121}}{\left({x}+\mathrm{6}\right)^{\mathrm{2}} +\mathrm{25}}\:=\:\frac{\left({x}−\mathrm{6}\right)^{\mathrm{2}} }{\left({x}+\mathrm{6}\right)^{\mathrm{2}} } \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{36}\right)^{\mathrm{2}} +\mathrm{121}\left({x}+\mathrm{6}\right)^{\mathrm{2}} =\left({x}^{\mathrm{2}} −\mathrm{36}\right)^{\mathrm{2}} +\mathrm{25}\left({x}−\mathrm{6}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{11}{x}+\mathrm{66}\right)^{\mathrm{2}} =\left(\mathrm{5}{x}−\mathrm{30}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{16}{x}+\mathrm{36}\right)\left(\mathrm{6}{x}+\mathrm{96}\right)=\mathrm{0} \\ $$$$\:\begin{cases}{{x}=−\frac{\mathrm{9}}{\mathrm{4}}}\\{{x}=−\mathrm{16}}\end{cases} \\ $$$${for}\:{x}=−\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${y}=\sqrt{\left(−\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{6}\right)^{\mathrm{2}} +\mathrm{25}}\:+\sqrt{\left(−\frac{\mathrm{9}}{\mathrm{4}}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{121}} \\ $$$${y}=\sqrt{\frac{\mathrm{625}}{\mathrm{16}}}\:+\:\sqrt{\frac{\mathrm{3025}}{\mathrm{16}}}\:=\:\frac{\mathrm{25}}{\mathrm{4}}+\frac{\mathrm{55}}{\mathrm{4}}=\mathrm{20} \\ $$$${for}\:{x}=−\mathrm{16} \\ $$$${y}=\sqrt{\left(−\mathrm{16}+\mathrm{6}\right)^{\mathrm{2}} +\mathrm{25}}\:+\sqrt{\left(−\mathrm{16}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{121}} \\ $$$${y}=\sqrt{\mathrm{125}}\:+\sqrt{\mathrm{605}}\:=\:\mathrm{35}.\mathrm{78} \\ $$$${therefore}\:{minimum}\:{value}\:{is}\:\mathrm{20} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *