1961 AHSME Problems/Problem 40

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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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