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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 MM be the maximum value of the expression
A=x4y+x3y+x2y+xy+xy2+xy3+xy4,A=x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4,
 subject to x+y=3x+y=3, where xx and yy are real numbers. Find the greatest integer not exceeding MM.
Solution
Answer: 36
Step 1: Pair the symmetric terms and take out the common factor xyxy from each pair:
A=(x4y+xy4)+(x3y+xy3)+(x2y+xy2)+xyA=(x^4y+xy^4)+(x^3y+xy^3)+(x^2y+xy^2)+xy
=xy(x3+y3)+xy(x2+y2)+xy(x+y)+xy.=xy(x^3+y^3)+xy(x^2+y^2)+xy(x+y)+xy.
 Step 2: Express the symmetric sums through x+yx+y and xyxy, using the standard identities x3+y3=(x+y)3−3xy(x+y)x^3+y^3=(x+y)^3-3xy(x+y) and x2+y2=(x+y)2−2xyx^2+y^2=(x+y)^2-2xy:
A=xy[(x+y)3−3xy(x+y)]+xy[(x+y)2−2xy]+xy(x+y)+xy.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=3x+y=3:
A=xy(27−9xy)+xy(9−2xy)+3xy+xy.A=xy(27-9xy)+xy(9-2xy)+3xy+xy.
 Step 4: Let t=xyt=xy and expand:
A=(27t−9t2)+(9t−2t2)+3t+t=40t−11t2.A=(27t-9t^2)+(9t-2t^2)+3t+t=40t-11t^2.
 Step 5: Complete the square in tt:
A=−11t2+40t=−11(t−2011)2+40011≤40011.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=2011t=xy=\dfrac{20}{11}. Step 6: Check that this value of tt is attainable. With x+y=3x+y=3 and xy=2011xy=\dfrac{20}{11}, the numbers x,yx,y are roots of u2−3u+2011=0u^2-3u+\dfrac{20}{11}=0, whose discriminant is 9−4⋅2011=1911>09-4\cdot\dfrac{20}{11}=\dfrac{19}{11}>0. So real x,yx,y exist, and the maximum is achieved. Step 7: Therefore
M=40011≈36.36,so [M]=36.M=\frac{400}{11}\approx 36.36,\qquad\text{so } [M]=36.
 Answer: 36
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