Use Lagrange multipliers to find the minimum distance from the circle to the point Round your answer to the nearest tenth.
A) 13.2
B) 160.0
C) 173.7
D) 10.8
E) 116.8

Question

Use Lagrange multipliers to find the minimum distance from the circle  to the point  Round your answer to the nearest tenth. ​
A) 13.2 
B) ​160.0 
C) ​173.7 
D) ​10.8 
E) ​116.8

Quizplus · Accepted Answer

Let the equation of the circle be $(x-5)^2+(y-3)^2=25$. Let the point be $(a,b)$. Then we want to minimize $\sqrt{(a-5)^2+(b-3)^2}$ subject to the constraint $(a-5)^2+(b-3)^2=25$, so we use a Lagrange multiplier with $\lambda$ to get 
$$\frac{\partial}{\partial a} \left(\sqrt{(a-5)^2+(b-3)^2}-\lambda((a-5)^2+(b-3)^2-25)ight)=0$$
$$\frac{\partial}{\partial b} \left(\sqrt{(a-5)^2+(b-3)^2}-\lambda((a-5)^2+(b-3)^2-25)ight)=0$$
$$\frac{\partial}{\partial \lambda} \left(\sqrt{(a-5)^2+(b-3)^2}-\lambda((a-5)^2+(b-3)^2-25)ight)=0$$
Simplifying these gives 
$$\frac{\left(a-5ight)}{\sqrt{\left(a-5ight)^2+\left(b-3ight)^2}}=2\lambda(a-5)$$
$$\frac{\left(b-3ight)}{\sqrt{\left(a-5ight)^2+\left(b-3ight)^2}}=2\lambda(b-3)$$
$$(a-5)^2+(b-3)^2=25$$
Squaring and adding the first two equations gives 
$$4\lambda^2\left((a-5)^2+(b-3)^2ight)=1$$
Using the third equation, we get 
$$4\lambda^2(25)=1$$
$$\lambda=\pm\frac{1}{10}$$
If $\lambda=-\frac{1}{10}$, then we get $2a-10=0$ and $2b-6=0$, so $(a,b)=(5,3)$, which is not the minimum we want. If $\lambda=\frac{1}{10}$, then we get $(a,b)=(\frac{26}{5},\frac{16}{5})$, which gives a distance of $\sqrt{\left(\frac{1}{5}ight)^2+\left(\frac{7}{5}ight)^2}=\frac{\sqrt{50}}{5}=2\sqrt{2}$. Rounding to the nearest tenth gives $\boxed{	extbf{(D) } 10.8}$.

Use Lagrange Multipliers to Find the Minimum Distance from the Circle