Difference between revisions of "1957 AHSME Problems/Problem 36"

(Created page with "By AM-GM, we have <cmath>\frac{x+y}{2} \geq \sqrt{xy}</cmath> Substituting, we have <cmath>\frac{1}{2} \geq \sqrt {xy}</cmath> <cmath>\frac{1}{4} \geq xy</cmath> Equality occu...")
 
(solution edits, see also box, statement of problem)
 
Line 1: Line 1:
By AM-GM, we have
+
== Problem ==
 +
If <math>x + y = 1</math>, then the largest value of <math>xy</math> is:
 +
 
 +
<math>\textbf{(A)}\ 1\qquad \textbf{(B)}\ 0.5\qquad \textbf{(C)}\ \text{an irrational number about }{0.4}\qquad \textbf{(D)}\ 0.25\qquad\textbf{(E)}\ 0</math>
 +
 
 +
== Solution ==
 +
By [[AM-GM]], we have
 
<cmath>\frac{x+y}{2} \geq \sqrt{xy}</cmath>
 
<cmath>\frac{x+y}{2} \geq \sqrt{xy}</cmath>
 
Substituting, we have
 
Substituting, we have
 
<cmath>\frac{1}{2} \geq \sqrt {xy}</cmath>
 
<cmath>\frac{1}{2} \geq \sqrt {xy}</cmath>
 
<cmath>\frac{1}{4} \geq xy</cmath>
 
<cmath>\frac{1}{4} \geq xy</cmath>
Equality occurs when <math>x = y = \frac{1}{2}</math>
+
Equality occurs when <math>x = y = \boxed{\textbf{(D) }\frac12}</math>.
<math>\boxed{D}</math>
 
  
 
~JustinLee2017
 
~JustinLee2017
 +
 +
== See Also ==
 +
{{AHSME 50p box|year=1957|num-b=35|num-a=37}}
 +
{{MAA Notice}}
 +
[[Category:AHSME]][[Category:AHSME Problems]]

Latest revision as of 09:13, 26 July 2024

Problem

If $x + y = 1$, then the largest value of $xy$ is:

$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 0.5\qquad \textbf{(C)}\ \text{an irrational number about }{0.4}\qquad \textbf{(D)}\ 0.25\qquad\textbf{(E)}\ 0$

Solution

By AM-GM, we have \[\frac{x+y}{2} \geq \sqrt{xy}\] Substituting, we have \[\frac{1}{2} \geq \sqrt {xy}\] \[\frac{1}{4} \geq xy\] Equality occurs when $x = y = \boxed{\textbf{(D) }\frac12}$.

~JustinLee2017

See Also

1957 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 35
Followed by
Problem 37
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 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

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