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 | ||
+ | |||
+ | <math> \textbf{(A)}\ \frac{1}{32}\qquad | ||
+ | \textbf{(B)}\ \frac{1}{36}\qquad | ||
+ | \textbf{(C)}\ \frac{4}{125}\qquad | ||
+ | \textbf{(D)}\ \frac{1}{127}\qquad | ||
+ | \textbf{(E)}\ \text{none of these} </math> | ||
− | |||
==Solution== | ==Solution== | ||
− | The | + | Suppose that the product <math>xyz</math> is minimized at <math>(x,y,z)=(x_0,y_0,z_0).</math> Without the loss of generality, let <math>x_0 \leq y_0 \leq z_0</math> and fix <math>y=y_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> 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 | ||
+ | |||
==See Also== | ==See Also== | ||
{{AHSME box|year=1982|num-b=28|num-a=30}} | {{AHSME box|year=1982|num-b=28|num-a=30}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 05:27, 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.