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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

The number of four-letter words that can be formed using the letters of the word $BARRACK$ is

The total number of three-digit numbers,divisible by $3$,which can be formed using the digits $1, 3, 5, 8$,if repetition of digits is allowed,is:

Five balls of different colours are to be placed in three boxes of different sizes. The number of ways in which we can place the balls in the boxes so that no box remains empty is

There are $3$ sections in a question paper and each section contains $5$ questions. $A$ candidate has to answer a total of $5$ questions,choosing at least one question from each section. Then the number of ways,in which the candidate can choose the questions,is

If the number of all possible permutations of the letters of the word $MATHEMATICS$ in which the repeated letters are not together is $90(X)$,then $X=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module