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Difference between revisions of "1961 AHSME Problems/Problem 40"

(Created page with "==Problem 40== Find the minimum value of <math>\sqrt{x^2+y^2}</math> if <math>5x+12y=60</math>. <math> \textbf{(A)}\ \frac{60}{13} \qquad\textbf{(B)}\ \frac{13}{5} \qquad\text...")
 
m (→‎Solution 3)
 
(14 intermediate revisions by 5 users not shown)
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==Problem 40==
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== Problem ==
  
 
Find the minimum value of <math>\sqrt{x^2+y^2}</math> if <math>5x+12y=60</math>.  
 
Find the minimum value of <math>\sqrt{x^2+y^2}</math> if <math>5x+12y=60</math>.  
  
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<math> \textbf{(A)}\ \frac{60}{13} \qquad\textbf{(B)}\ \frac{13}{5} \qquad\textbf{(C)}\ \frac{13}{12} \qquad\textbf{(D)}\ 1}\qquad\textbf{(E)}\ 0 } </math>
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<math>\textbf{(A)}\ \frac{60}{13}\qquad
 +
\textbf{(B)}\ \frac{13}{5}\qquad
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\textbf{(C)}\ \frac{13}{12}\qquad
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\textbf{(D)}\ 1\qquad
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\textbf{(E)}\ 0 </math>
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 +
==Solutions==
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===Solution 1===
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Let <math>x^2 + y^2 = r^2</math>, so <math>r = \sqrt{x^2 + y^2}</math>.  Thus, this problem is really finding the shortest distance from the origin to the line <math>5x + 12y = 60</math>.
 +
 
 +
<asy>import graph; size(8.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=13.2,ymin=-5.2,ymax=6.2;
 +
pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0);
 +
 
 +
/*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1;
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for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs);
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Label laxis; laxis.p=fontsize(10);
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xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
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draw((0,5)--(12,0),Arrows);
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draw(Circle((0,0),60/13),dotted);
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</asy>
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 +
From the graph, the shortest distance from the origin to the line is the [[altitude]] to the hypotenuse of the right triangle with legs <math>5</math> and <math>12</math>.  The hypotenuse is <math>13</math> and the area is <math>30</math>, so the altitude to the hypotenuse is <math>\frac{60}{13}</math>, which is answer choice <math>\boxed{\textbf{(A)}}</math>.
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===Solution 2===
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Solve for <math>y</math> in the linear equation.
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<cmath>12y = 60 - 5x</cmath>
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<cmath>y = 5 - \frac{5x}{12}</cmath>
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Substitute <math>y</math> in <math>\sqrt{x^2+y^2}</math>.
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<cmath>\sqrt{x^2 + (5 - \frac{5x}{12})^2}</cmath>
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<cmath>\sqrt{x^2 + 25 - \frac{25x}{6} + \frac{25x^2}{144}}</cmath>
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<cmath>\sqrt{\frac{169x^2}{144} - \frac{25x}{6} + 25}</cmath>
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To find the minimum, find the vertex of the quadratic.  The x-value of the vertex is <math>\frac{25}{6} \cdot \frac{72}{169} = \frac{300}{169}</math>.  Thus, the minimum value is
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<cmath>\sqrt{\frac{169}{144} \cdot \frac{300^2}{169^2} - \frac{25}{6} \cdot \frac{300}{169} + 25}</cmath>
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<cmath>\sqrt{\frac{10000}{16 \cdot 169} - \frac{1250}{169} + 25}</cmath>
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<cmath>\sqrt{\frac{625}{169} - \frac{1250}{169} + \frac{4225}{169}}</cmath>
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<cmath>\sqrt{\frac{3600}{169}}</cmath>
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<cmath>\frac{60}{13}</cmath>
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The answer is <math>\boxed{\textbf{(A)}}</math>.
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===Solution 3===
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By Cauchy-Schwarz,
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<cmath>(5^2 + 12^2)(x^2 + y^2) \geq (5x+12y)^2</cmath>
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Therefore:
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<cmath>(13^2)(x^2 + y^2) \geq 60^2</cmath>
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<cmath>13\sqrt{x^2 + y^2} \geq 60</cmath>
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Therefore:
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<cmath>\sqrt{x^2 + y^2} \geq \frac{60}{13}</cmath>
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Thus the answer is <math>\boxed{\textbf{(A)}}</math>.
 +
 
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==See Also==
 +
{{AHSME 40p box|year=1961|num-b=39|after=Last Question}}
 +
 
 +
[[Category:Intermediate Algebra Problems]]
 +
 
 +
{{MAA Notice}}

Latest revision as of 21:31, 11 June 2020

Problem

Find the minimum value of $\sqrt{x^2+y^2}$ if $5x+12y=60$.

$\textbf{(A)}\ \frac{60}{13}\qquad \textbf{(B)}\ \frac{13}{5}\qquad \textbf{(C)}\ \frac{13}{12}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$

Solutions

Solution 1

Let $x^2 + y^2 = r^2$, so $r = \sqrt{x^2 + y^2}$. Thus, this problem is really finding the shortest distance from the origin to the line $5x + 12y = 60$.

[asy]import graph; size(8.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=13.2,ymin=-5.2,ymax=6.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); draw((0,5)--(12,0),Arrows); draw(Circle((0,0),60/13),dotted); [/asy]

From the graph, the shortest distance from the origin to the line is the altitude to the hypotenuse of the right triangle with legs $5$ and $12$. The hypotenuse is $13$ and the area is $30$, so the altitude to the hypotenuse is $\frac{60}{13}$, which is answer choice $\boxed{\textbf{(A)}}$.

Solution 2

Solve for $y$ in the linear equation. \[12y = 60 - 5x\] \[y = 5 - \frac{5x}{12}\] Substitute $y$ in $\sqrt{x^2+y^2}$. \[\sqrt{x^2 + (5 - \frac{5x}{12})^2}\] \[\sqrt{x^2 + 25 - \frac{25x}{6} + \frac{25x^2}{144}}\] \[\sqrt{\frac{169x^2}{144} - \frac{25x}{6} + 25}\] To find the minimum, find the vertex of the quadratic. The x-value of the vertex is $\frac{25}{6} \cdot \frac{72}{169} = \frac{300}{169}$. Thus, the minimum value is \[\sqrt{\frac{169}{144} \cdot \frac{300^2}{169^2} - \frac{25}{6} \cdot \frac{300}{169} + 25}\] \[\sqrt{\frac{10000}{16 \cdot 169} - \frac{1250}{169} + 25}\] \[\sqrt{\frac{625}{169} - \frac{1250}{169} + \frac{4225}{169}}\] \[\sqrt{\frac{3600}{169}}\] \[\frac{60}{13}\] The answer is $\boxed{\textbf{(A)}}$.

Solution 3

By Cauchy-Schwarz, \[(5^2 + 12^2)(x^2 + y^2) \geq (5x+12y)^2\] Therefore: \[(13^2)(x^2 + y^2) \geq 60^2\] \[13\sqrt{x^2 + y^2} \geq 60\] Therefore: \[\sqrt{x^2 + y^2} \geq \frac{60}{13}\] Thus the answer is $\boxed{\textbf{(A)}}$.

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 39 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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