Maximizing the w

Algebra Level 5

Suppose w , x , y , z w,x,y,z satisfy w + x + y + z = 25 , w x + w y + w z + x y + x z + y z = 2 y + 2 z + 193 \begin{aligned}w+x+y+z&=25,\\wx+wy+wz+xy+xz+yz&=2y+2z+193\end{aligned} The largest possible value of w w can be expressed in lowest terms as w 1 / w 2 w_1/w_2 for some integers w 1 , w 2 > 0 w_1,w_2>0 . Find w 1 + w 2 w_1+w_2 .


The answer is 27.

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Alan Yan by Alan Yan , 20, USA

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1 solution

Alan Yan
Oct 20, 2015

625 = w 2 + x 2 + y 2 + z 2 + 2 w x + 2 w y + 2 w z + 2 x y + 2 x z + 2 y z = w 2 + y 2 + z 2 + x 2 + 4 y + 4 z + 386 w 2 + x 2 + ( y + 2 ) 2 + ( z + 2 ) 2 = 247 w + x + ( y + 2 ) + ( z + 2 ) = 29 ( 1 + 1 + 1 ) ( x 2 + ( y + 2 ) 2 + ( z + 2 ) 2 ) ≥ ( x + ( y + 2 ) + ( z + 2 ) ) 2 3 ( 247 − w 2 ) ≥ ( 29 − w ) 2 2 w 2 − 29 w + 50 ≤ 0 ( 2 w − 25 ) ( w − 2 ) ≤ 0 w ∈ [ 2 , 25 2 ] \begin{aligned} 625 = w^2 + x^2 + y^2 + z^2 + 2wx + 2wy + 2wz + 2xy + 2xz + 2yz & = w^2 + y^2 + z^2 + x^2 + 4y + 4z + 386\\ w^2 + x^2 + (y + 2)^2 + (z + 2)^2 & = 247 \\ w + x + (y +2) + (z+2) & = 29 \\ (1 + 1 + 1)(x^2 + (y+2)^2 + (z+2)^2) & \geq (x + (y+2) + (z+2))^2 \\ 3(247 - w^2) & \geq (29 - w)^2 \\ 2w^2 - 29w + 50 & \leq 0 \\ (2w - 25)(w - 2) & \leq 0 \\ w & \in \left[2 , \frac{25}{2}\right] \end{aligned}

Thus the answer is 25 + 2 = 27 . 25 + 2 = \boxed{27}.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

I salute you!

Adarsh Kumar - 5 years, 7 months ago

Awesome solution!

Harsh Shrivastava - 5 years, 7 months ago

Can you explain from where you got that 3rd line in the equation?

Edgar Wang - 5 years, 7 months ago

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w + x + y + z = 25 ⟹ w + x + ( y + 2 ) + ( z + 2 ) = 25 + 4 = 29 w +x + y +z = 25 \implies w + x + (y+2) + (z+2) = 25 + 4 = 29

Alan Yan - 5 years, 7 months ago

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oops how did I miss that :P

Edgar Wang - 5 years, 7 months ago

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