I have a word problem

Question

Find 2 numbers such that one number is 1 unit greater than the second number and the sum of the squares of the numbers is a minimum.

Answer 1

Find 2 numbers such that one number is 1 unit greater than the second number and the sum of the squares of the numbers is a minimum.

x = 2nd number

y = 1st number

y = x + 1

S = x^2 + y^2 = x^2 + (x + 1)^2

S = x^2 + x^2 + 2x + 1

S = 2x^2 + 2x + 1

The minimum can be found with algebra or calculus.

A. Algebra solution:

Parabola opening up, so Vertex is minimum point:

a(x-h)^2+k =
ax^2 – 2ahx + ah^2 + k
S = 2x^2 + 2x + 1
a = 2
– 2ah = 2, h = 2/(–2(2)) = – 1/2
ah^2 + k = 1, k = 1 – 2(–1/2)^2 = 1/2

Vertex = (–1/2,1/2)

The two numbers are:

x = –1/2

y = x + 1 = 1/2.

C. Calculus solution:

S = 2x^2 + 2x + 1

Extrema occur when S’ = 0:

S’ = 4x + 2 = 0

x = –1/2

If curvature is positive, S’’ > 0, then extremum will be a minimum:

S’’ = 4 > 0 ==> x = –1/2 locates a minimum of S.

y = x + 1 = 1/2.

Answer 2

Y = X +1

 

Y^2 + X^2  is minimum

 

  (X+1 )^2 + X^2

 

  2X^2 +2X + 1

 

  2 ( X +1/2 ) ^2 + 1 =

 

   2 ( X +1/2 ) ^2  + 1/2

 

   (X+1/2 ) ^2 ≥0

 

   Minimum value  X :   X + 1/2 =0     X = -1/2

 

 

 

Answer 3

Hi Milton,

Let one number be x, then the other is x+1, so the sum of the squares of the two numbers is x²+(x+1)²=2(x²+x)+1=2(x+1/2)²+1/2. Based on the property of 2(x+1/2)²+1/2, when x=-1/2, 2(x+1/2)²+1/2 achieves the minimum 1/2. So x=-1/2, x+1=1/2, which means the 2 numbers are -1/2, 1/2.

 

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