You gotta think of it in some other way
If 8 x + 1 5 y = 1 2 0 , the minimum value of x 2 + y 2 can be expressed as b a where a and b are coprime positive integers, find a + b .
The answer is 137.
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4 solutions
did it the same way
Using Cauchy Schwarz inequality, we have:
( 8 x + 1 5 y ) 2 1 2 0 2 x 2 + y 2 ⇒ x 2 + y 2 ⇒ a + b ≤ ( 8 2 + 1 5 2 ) ( x 2 + y 2 ) ≤ 1 7 2 ( x 2 + y 2 ) ≥ 1 7 2 1 2 0 2 ≥ 1 7 1 2 0 = 1 2 0 + 1 7 = 1 3 7
Nice solution. Very succinct. +1
Extension: Generalise the minimum value of x 2 + y 2 for any equation of a line i.e. a x + b y = c with proof.
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If a x + b y = c , then using Cauchy Schwarz inequality, we have:
( a x + b y ) 2 c 2 x 2 + y 2 ⇒ x 2 + y 2 ≤ ( a 2 + b 2 ) ( x 2 + y 2 ) ≤ ( a 2 + b 2 ) ( x 2 + y 2 ) ≥ a 2 + b 2 c 2 ≥ a 2 + b 2 c
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What if , if we take x=15cos^2(A) and y=8sin^2(A) . Then by RMS>=AM . Minimum value comes out to be 23/sqrt2
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@Aakash Khandelwal – But why x = 1 5 cos 2 θ and y = 6 sin 2 θ ?
(Old style calculus way)
A very simple way to look at it would be to consider two vectors a=8i+15j and b=xi+yj.
Now, a.b=|a||b|Cos(ω). => |b| = 120/17 sec ω. For minimum case, sec ω =1. Hence the answer is 120+17=137.
8 x + 1 5 y = 1 2 0 represents a straight line. x 2 + y 2 represents the distance of any point on that line from the origin. To minimize it, we take the perpendicular distance that is 8 2 + 1 5 2 1 2 0 = 1 7 1 2 0
∴ a + b = 1 3 7