Difference between revisions of "1961 AHSME Problems/Problem 40"
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− | ==Problem 40== | + | == 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> | + | <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> | ||
+ | |||
+ | ==Solutions (WIP)== | ||
+ | |||
+ | ===Solution 1=== | ||
+ | Solve for <math>y</math> in the linear equation. | ||
+ | <cmath>12y = 60 - 5x</cmath> | ||
+ | <cmath>y = 5 - \frac{5x}{12}</cmath> | ||
+ | Substitute <math>y</math> in <math>\sqrt{x^2+y^2}</math>. | ||
+ | <cmath>\sqrt{x^2 + (5 - \frac{5x}{12})^2}</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> | ||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME 40p box|year=1961|num-b=26|num-a=28}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:56, 25 May 2018
Problem 40
Find the minimum value of if .
Solutions (WIP)
Solution 1
Solve for in the linear equation. Substitute in .
Solution 2
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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All AHSME Problems and Solutions |
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