Question

Let L be the line with equation y=x-1. Find the minimum distance between L and the point (4,5) by using (a) theorem 10.2.4, (b) the Method of Lagrange.

Answer 1

$$d^2=(x-4)^2+(y-5)^2$$

(a) Given $y=x-1.$

Then

$$d^2=f(x)=(x-4)^2+(x-1-5)^2$$

$$f'(x)=2(x-4)+2(x-6)$$

$$=2(x-4+x-6)=4(x-5)$$

Find the critical number(s):

$$f'(x)=0=>4(x-5)=0=>x=5$$

Critical number: $5.$

$$f''(x)=4>0$$

Then the function $f(x)$ has a local minimum at $x=4.$

Since the function $f$ has the only extremum, then the function $f(x)$ has an absolute minimum at $x=4.$

$$x=5, y=5-1=4$$

$$d=\sqrt{(5-4)^2+(4-5)^2}=\sqrt{2}$$

(b) Given $y=x-1.$

Then

$$L(x,y)=(x-4)^2+(y-5)^2-\lambda(x-y-1)$$

$$\dfrac{\partial L}{\partial x}=2(x-4)-\lambda$$

$$\dfrac{\partial L}{\partial y}=2(y-5)+\lambda$$

$$\dfrac{\partial L}{\partial \lambda}=-x+y+1$$

$$\dfrac{\partial L}{\partial x}=0$$

$$\dfrac{\partial L}{\partial y}=0$$

$$\dfrac{\partial L}{\partial \lambda}=0$$

$$\lambda=2(x-4)$$

$$y=x-1$$

$$2(x-1-5)+2(x-4)=0$$

$$x=5$$

$$y=4$$

$$\lambda=2$$

The minimum distance between L and the point (4,5) is $\sqrt{2}.$