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If w, x, y, and z are positive real numbers such that w + 2x + 3y + 4z = 8 then what is the maximum value of wxyz?

 
 Oct 11, 2021
 #1
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The minimum value is 3/4.

 
 Oct 11, 2021
 #2
avatar+114826 
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Original was deleted because it had a silly error

 

w + 2x + 3y + 4z = 8

If they are all the same, say A

then

A+2A+3A+4A=8

10A=8

A=8/10

 

A^4 = .8^4 = 0.4096    Maybe

 
 Oct 12, 2021
edited by Melody  Oct 12, 2021
edited by Melody  Oct 12, 2021
edited by Melody  Oct 12, 2021
 #3
avatar+11 
+2

The correct answer is 2/3 because you get (w+2x+3y+4z)/4 >= 4rt(24wxyz) which becomes 8/4 >= 4rt(24wxyz) so you get 2^4 >= 24wxyz. This simplifies to 16/24=2/3=wxyz.

 
 Oct 12, 2021
edited by Guest  Oct 12, 2021
 #5
avatar+114826 
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This explanation makes no sense to me.

 
Melody  Oct 12, 2021
 #6
avatar+11 
+1

You use the AM-GM inequality which says (a_1+a_2+a_3+...+a_n)/n >= nrt(a_1*a_2*...*a_n). Here's the art of problem solving article https://artofproblemsolving.com/wiki/index.php/Arithmetic_Mean-Geometric_Mean_Inequality

 
Tacoeggegg  Oct 12, 2021
 #7
avatar+114826 
0

ok thanks, I will look at it properly later on    smiley

 
Melody  Oct 12, 2021
 #9
avatar+114826 
0

I agree my answer is not correct but there is not enough information on that linked page for your answer to be reached.

 

You should not be dividing by 4 you should be dividing by 10. (to get the arithemtic mean)

 

the numbers are  w,x,x,y,y,y,z,z,z,z,    There are 10 numbers.

 

Perhaps you would like to provide more working or another link address that you used.

 
Melody  Oct 14, 2021

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