If a + b + c = 50 and a, b, c are non-negativ | vedclass

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(A) We want to maximize $f(a, b, c) = ab^2c$ subject to $a + b + c = 50$,where $a, b, c$ are non-negative even integers.Let $a = 2x, b = 2y, c = 2z$.

Vedclass · Accepted Answer

$97344$

Vedclass · Answer

(A) We want to maximize $f(a, b, c) = ab^2c$ subject to $a + b + c = 50$,where $a, b, c$ are non-negative even integers.Let $a = 2x, b = 2y, c = 2z$. Then $2x + 2y + 2z = 50$,which implies $x + y + z = 25$.The expression becomes $f = (2x)(2y)^2(2z) = 16xy^2z$.To maximize $xy^2z$ subject to $x + y + z = 25$,we use the $AM$-$GM$ inequality.We can write $xy^2z = 4 	imes x 	imes (\frac{y}{2}) 	imes (\frac{y}{2}) 	imes z$.The sum of these terms is $x + \frac{y}{2} + \frac{y}{2} + z = x + y + z = 25$.By $AM$-$GM$,the product is maximized when $x = \frac{y}{2} = z$.Substituting this into the sum: $x + 2x + x = 25 \implies 4x = 25 \implies x = 6.25$.Since $a, b, c$ must be even integers,$x, y, z$ must be integers.We test values near $x = 6.25$ and $y = 12.5$.If $y = 12$,then $x + z = 13$. For $x=z$,$x=6.5$ (not integer). Try $x=6, z=7$ or $x=7, z=6$.If $y = 12, x = 6, z = 7$,then $a = 12, b = 24, c = 14$. $ab^2c = 12 	imes 24^2 	imes 14 = 96768$.If $y = 14$,then $x + z = 11$. Try $x=5, z=6$. $a = 10, b = 28, c = 12$. $ab^2c = 10 	imes 28^2 	imes 12 = 94080$.If $y = 12, x = 6.5$ is not possible,try $b=26$. Then $a+c = 24$. For $a=c=12$,$ab^2c = 12 	imes 26^2 	imes 12 = 97344$.Comparing these,the maximum value is $97344$.

If $a + b + c = 50$ and $a, b, c$ are non-negative even integers,then the greatest value of $ab^2c$ is:

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