Question 2

Find the point on the graph of y = x² that is the smallest distance from the point (0, 4).



Solution

Answer at point x= +√(7/2), with a value of 2.8343.

The question is an optimization question however we are looking for a minimum value so it is really called a minimization question.

∙ Like other optimization problems you need to find what is always true using geometry or an equation, and identify your constraint.
In this case you will be using the distance formula D = √(y² + x²). And your constraint is y = x². However, square roots may be difficult to work with so you'd want to square both sides of the distance formula to get
D² = (y²-4)² + x²

∙ And substituting y for x²
D² = (x²-4)² + x²

∙ Find the derivative of the equation in order to find its roots.
D²'(x) = 2(x²-4)(2x) + 2x
D²'(x) = 4x³-16x + 2x
D²'(x) = 2x(2x²-7)
D²'(x) = 0 at x = ±√(7/2) and x=0

∙ Using a number line we can pinpoint where a minimum is located.
D²'(x) = ___-______+_____-_____+___
-√(7/2) 0 +√(7/2)

∙ A minimum occurs when the derivative changes from negative to positive.
In this case it happens at x= -√(7/2) and x= +√(7/2). So we would have to investigates these two values.

∙ So going back to using our original distance formula equation and plugging in the values
D(x) = √((x-4)²) + x²)
D(-√(7/2)) = √((-√(7/2))-4)²) - √(7/2)²) = 6.1617
D(+√(7/2)) = √((+√(7/2))-4)²) + √(7/2)²) = 2.8343

∙ Therefore the minimum value is at x = +√(7/2).