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

(Problem 40)
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==Problem 40==
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== Problem 40==
  
 
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>.  
  
<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
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\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 (WIP)==
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===Solution 1===
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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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===Solution 2===
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==See Also==
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{{AHSME 40p box|year=1961|num-b=26|num-a=28}}
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[[Category:Intermediate Algebra Problems]]
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:56, 25 May 2018

Problem 40

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 (WIP)

Solution 1

Solve for $y$ in the linear equation. \[12y = 60 - 5x\] \[y = 5 - \frac{5x}{12}\] Substitute $y$ in $\sqrt{x^2+y^2}$. \[\sqrt{x^2 + (5 - \frac{5x}{12})^2}\] \[\sqrt{x^2 + 25 - \frac{25x}{6} + \frac{25x^2}{144}}\] \[\sqrt{\frac{169x^2}{144} - \frac{25x}{6} + 25}\]

Solution 2

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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