If x,y,z are positive integer and x+y+z =21, then find the maximum value for x^2 * y^3 * z^2

944784{ x=6, y=9, z=6}

Some types of question should be solved using logic like aptitude .<br>Take x=y=z which will be 49*343*49 = 2401 * 343 = 823543<br><br>Now increase the value of y (as it contains cube) by 2 and decrease x & z 1 each <br>X 6 , y 9, z 6 =36*729*36 = 944784 <br> Check once again if increasing y by 2 will increase the total or not .<br><br>Use calculator or speed math<br>

i tried weighted AM&gt;=GM, we can write x+y+z = (2x/2)+(3y/3)+(2z/2) = (x/2)+(x/2) + (y/3) + (y/3)+(y/3)+ (z/2)+ (z/2) = 21, the division is as we want to obtain the power of x,y,z as reqd. so (x+y+z)/7 &gt;= ((y/3)^3 * (x/2)^2 * (z/2)^2)^(1/7)=((y^3*x^2*z^2) * 1/(27*4*4))^(1/7)... Now taking 7th power on either sides of the inequality we get max value of the given expn. = 27*16*21^7/(7^7)=944784

2^4*3^1o

Please explain the method to derive the answer too