Graphing/Optimization

Question

Find the point on the graph of y=x² +1 that is closest to the point (8, 3/2).

Answer 1

A. Solution by distance:

Find the distance, d, between the point (x,x^2+1) and (8,3/2):

d^2 = (x–8)^2 + (x^2+1–3/2)^2

= x^2–16x+64 + (x^2–1/2)^2

= x^2–16x+64 + x^4–x^2+1/4

= –16x+64 + x^4+1/4

4d^2 = 4x^4 – 64x + 257 = f(x)

d will be minimum when f is minimum.

f has extrema when f'(x) = 0:

f'(x) = 16x^3 – 64 = 0

16x^3 = 64

x^3 = 4

x = 4^(1/3) = 2^(2/3)

y = (2^(2/3))^2 + 1 = 2^(4/3) + 1

The point on the graph of y=x^2 +1 that is closest to the point (8, 3/2) is:

(2^(2/3),2^(4/3) + 1) ≈ (1.587401051968203, 3.51984209978976)

B. Solution by Normal Line:

Find the Line Normal to y = x^2 + 1 that goes through (8,3/2):

y’ = 2x

Normal “Line”:

y – 3/2 = –1/(2x) (x – 8) = –1/2 + 4/x

y = 1 + 4/x

Answer will be intersection of Normal “Line” and y = x^2 + 1.

y = x^2 + 1 = 1 + 4/x

x^2 + 1 = 1 + 4/x

x^3 = 2^2

x = 2^(2/3)

y = (2^(2/3))^2 + 1 = 2^(4/3)) + 1