Minimum value of equation

Algebra Level 2

Given P = ( x + 3 y 5 ) 2 6 x y + 26 P=(x+3y-5)^{2}-6xy+26 . Find the minimum value of P P

Bonus: Which value of x , y x, y can achieve this?


The answer is 1.

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2 solutions

Chew-Seong Cheong
Jun 19, 2019

P = ( x + 3 y 5 ) 2 6 x y + 26 = x 2 + 9 y 2 + 6 x y 10 x 30 y + 25 6 x y + 26 = x 2 + 9 y 2 10 x 30 y + 51 = ( x 2 10 x + 25 ) + ( 9 y 2 30 y + 25 ) + 1 = ( x 5 ) 2 + ( 3 y 5 ) 2 + 1 \begin{aligned} P & = (x+3y-5)^2 - 6xy + 26 \\ & = x^2 + 9y^2 + 6xy - 10x - 30y + 25 - 6xy + 26 \\ & = x^2 + 9y^2 - 10x - 30y + 51 \\ & = (x^2 - 10x + 25) + (9y^2 - 30y + 25) + 1 \\ & = (x - 5)^2 + (3y - 5)^2 + 1 \end{aligned}

Therefore,

min ( P ) = min ( ( x 5 ) 2 + ( 3 y 5 ) 2 + 1 ) = min ( ( x 5 ) 2 ) + min ( ( 3 y 5 ) 2 ) + 1 = 0 + 0 + 1 = 1 when x = 5 , y = 5 3 \begin{aligned} \min (P) & = \min \left((x - 5)^2 + (3y - 5)^2 + 1\right) \\ & = \min \left((x - 5)^2 \right) + \min \left((3y - 5)^2 \right) + 1 \\ & = 0 + 0 + 1 = \boxed 1 & \small \color{#3D99F6} \text{when }x=5, y = \frac 53 \end{aligned}

Chris Lewis
Jun 19, 2019

There are no constraints, so partial differentiation is the easiest (but probably not the most elegant) way to solve this. We have

P x = 2 ( x + 3 y 5 ) 6 y = 2 ( x 5 ) P y = 6 ( x + 3 y 5 ) 6 x = 6 ( 3 y 5 ) P_x=2(x+3y-5)-6y=2(x-5)\\ P_y=6(x+3y-5)-6x=6(3y-5)

Setting both of these equal to zero we get x = 5 x=5 , y = 5 3 y=\frac53 and substituting in, P = 1 P=\boxed1 (it's easy to see from the form of P P that this must be a minimum).

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