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Maximum of x^4y+...+xy^4 given x+y=3: Find [M] | IOQM

JEE Maths question with a full step-by-step solution.

Question

Let

M

be the maximum value of the expression

A=x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4,

subject to

x+y=3

, where

x

and

y

are real numbers. Find the greatest integer not exceeding

M

Solution

Answer: 36

Step 1: Pair the symmetric terms and take out the common factor

xy

from each pair:

A=(x^4y+xy^4)+(x^3y+xy^3)+(x^2y+xy^2)+xy

=xy(x^3+y^3)+xy(x^2+y^2)+xy(x+y)+xy.

Step 2: Express the symmetric sums through

x+y

and

xy

, using the standard identities

x^3+y^3=(x+y)^3-3xy(x+y)

and

x^2+y^2=(x+y)^2-2xy

A=xy\big[(x+y)^3-3xy(x+y)\big]+xy\big[(x+y)^2-2xy\big]+xy(x+y)+xy.

Step 3: Substitute the constraint

x+y=3

A=xy(27-9xy)+xy(9-2xy)+3xy+xy.

Step 4: Let

t=xy

and expand:

A=(27t-9t^2)+(9t-2t^2)+3t+t=40t-11t^2.

Step 5: Complete the square in

t

A=-11t^2+40t=-11\left(t-\frac{20}{11}\right)^2+\frac{400}{11}\le\frac{400}{11}.

The maximum is reached when

t=xy=\dfrac{20}{11}

. Step 6: Check that this value of

t

is attainable. With

x+y=3

and

xy=\dfrac{20}{11}

, the numbers

x,y

are roots of

u^2-3u+\dfrac{20}{11}=0

, whose discriminant is

9-4\cdot\dfrac{20}{11}=\dfrac{19}{11}>0

. So real

x,y

exist, and the maximum is achieved. Step 7: Therefore

M=\frac{400}{11}\approx 36.36,\qquad\text{so } [M]=36.

Answer: 36

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