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

(Problem 40)
(Solution 1)
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<cmath>\sqrt{x^2 + 25 - \frac{25x}{6} + \frac{25x^2}{144}}</cmath>
 
<cmath>\sqrt{x^2 + 25 - \frac{25x}{6} + \frac{25x^2}{144}}</cmath>
 
<cmath>\sqrt{\frac{169x^2}{144} - \frac{25x}{6} + 25}</cmath>
 
<cmath>\sqrt{\frac{169x^2}{144} - \frac{25x}{6} + 25}</cmath>
 +
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
  
 
===Solution 2===
 
===Solution 2===

Revision as of 12:04, 29 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 $\sqrt{x^2+y^2}$. \[\sqrt{x^2 + (5 - \frac{5x}{12})^2}\] \[\sqrt{x^2 + 25 - \frac{25x}{6} + \frac{25x^2}{144}}\] \[\sqrt{\frac{169x^2}{144} - \frac{25x}{6} + 25}\] 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

Solution 2

See Also

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