Difference between revisions of "1982 AHSME Problems/Problem 29"
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==Problem== | ==Problem== | ||
− | Let <math>x,y</math> | + | Let <math>x,y,</math> and <math>z</math> be three positive real numbers whose sum is <math>1.</math> If no one of these numbers is more than twice any other, |
then the minimum possible value of the product <math>xyz</math> is | then the minimum possible value of the product <math>xyz</math> is | ||
Line 7: | Line 7: | ||
\textbf{(C)}\ \frac{4}{125}\qquad | \textbf{(C)}\ \frac{4}{125}\qquad | ||
\textbf{(D)}\ \frac{1}{127}\qquad | \textbf{(D)}\ \frac{1}{127}\qquad | ||
− | \textbf{(E)}\ \text{none of these} </math> | + | \textbf{(E)}\ \text{none of these} </math> |
==Solution== | ==Solution== | ||
Line 14: | Line 14: | ||
To minimize <math>xy_0z,</math> we minimize <math>xz.</math> Note that <math>x+z=1-y_0.</math> By a corollary of the AM-GM Inequality <i><b>(If two nonnegative numbers have a constant sum, then their product is minimized when they are as far as possible.)</b></i>, we get <math>z_0=2x_0.</math> It follows that <math>y_0=1-3x_0.</math> | To minimize <math>xy_0z,</math> we minimize <math>xz.</math> Note that <math>x+z=1-y_0.</math> By a corollary of the AM-GM Inequality <i><b>(If two nonnegative numbers have a constant sum, then their product is minimized when they are as far as possible.)</b></i>, we get <math>z_0=2x_0.</math> It follows that <math>y_0=1-3x_0.</math> | ||
− | Recall that <math>x_0 \leq 1-3x_0 \leq 2x_0,</math> so <math>\frac15 \leq x_0 \leq \frac14.</math> This problem is equivalent to finding the minimum value of <cmath>f(x)=xyz=x(1-3x)(2x)=2x^2(1-3x)</cmath> in the interval <math>I=\left[\frac15,\frac14\right].</math> | + | Recall that <math>x_0 \leq 1-3x_0 \leq 2x_0,</math> so <math>\frac15 \leq x_0 \leq \frac14.</math> This problem is equivalent to finding the minimum value of <cmath>f(x)=xyz=x(1-3x)(2x)=2x^2(1-3x)</cmath> in the interval <math>I=\left[\frac15,\frac14\right].</math> The graph of <math>y=f(x)</math> is shown below: |
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(900,200); | ||
+ | |||
+ | real f(real x) { return 2x^2 * (1-3x); } | ||
+ | |||
+ | real xMin = -0.349; | ||
+ | real xMax = 1/2; | ||
+ | real yMin = -1/4; | ||
+ | real yMax = 1/2; | ||
+ | |||
+ | draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); | ||
+ | draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); | ||
+ | label("$x$",(xMax,0),(2,0)); | ||
+ | label("$y$",(0,yMax),(0,2)); | ||
+ | |||
+ | pair A[]; | ||
+ | A[0] = (0,0); | ||
+ | A[1] = (1/2,1/2); | ||
+ | A[2] = (2/3,1/3); | ||
+ | |||
+ | draw((1/5,-0.015)--(1/5,0.015),linewidth(1)); | ||
+ | draw((1/4,-0.015)--(1/4,0.015),linewidth(1)); | ||
+ | draw((1/3,-0.015)--(1/3,0.015),linewidth(1)); | ||
+ | |||
+ | draw((1/5,yMin)--(1/5,yMax),dashed); | ||
+ | draw((1/4,yMin)--(1/4,yMax),dashed); | ||
+ | |||
+ | label("$0$",A[0],(-2,-2)); | ||
+ | label("$\frac15$",(1/5,0),(0,-2),UnFill); | ||
+ | label("$\frac14$",(1/4,0),(0,-2),UnFill); | ||
+ | label("$\frac13$",(1/3,0),(0,-2)); | ||
+ | |||
+ | draw(graph(f,xMin,xMax),red); | ||
+ | dot(A[0],red+linewidth(5)); | ||
+ | dot((1/3,0),red+linewidth(5)); | ||
+ | </asy> | ||
+ | Since <math>f</math> has a relative minimum at <math>x=0,</math> and cubic functions have at most one relative minimum, we conclude that the minimum value of <math>f</math> in <math>I</math> is at either <math>x=\frac15</math> or <math>x=\frac14.</math> As <math>f\left(\frac14\right)=\frac{1}{32}\leq f\left(\frac15\right)=\frac{4}{125},</math> the minimum value of <math>f</math> in <math>I</math> is <math>\boxed{\textbf{(A)}\ \frac{1}{32}}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM |
Latest revision as of 06:27, 18 September 2021
Problem
Let and
be three positive real numbers whose sum is
If no one of these numbers is more than twice any other,
then the minimum possible value of the product
is
Solution
Suppose that the product is minimized at
Without the loss of generality, let
and fix
To minimize we minimize
Note that
By a corollary of the AM-GM Inequality (If two nonnegative numbers have a constant sum, then their product is minimized when they are as far as possible.), we get
It follows that
Recall that so
This problem is equivalent to finding the minimum value of
in the interval
The graph of
is shown below:
Since
has a relative minimum at
and cubic functions have at most one relative minimum, we conclude that the minimum value of
in
is at either
or
As
the minimum value of
in
is
~MRENTHUSIASM
See Also
1982 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.