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

Revision as of 11:04, 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

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 19: / Line 19: @@
 <cmath>\sqrt{x^2 + 25 - \frac{25x}{6} + \frac{25x^2}{144}}</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===

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