How to Apply Lagrange multiplier well

Question

This problem was originally:

1. Sphere $S$: $x^2+y^2+z^2=50$
2. Point $P(0,5,5)$
3. Circle $C$ is an intersection of a plane that passes $P$, and sphere $S$.
4. The radius of the circle $C$ is $1$.

Find the maximum of area of the projection of $C$ on the $xy$ plane.

I want to apply lagrange multiplier on this problem. So I converted to:

Constraints: $$a^2+b^2+c^2=1$$ $$b+c=\frac 75$$

Problem: Find the minimum of $$c$$.

Where the vector $(a,b,c)$ is the normal of the plane that passes $P$.

Solution using Cauchy- Schwarz Inequality:

$$0 a+b=\frac 75 -c$$ $$a^2+b^2=1-c^2$$ $$(0^2+1^2)(a^2+b^2) \ge(0 a+b)^2$$ $$\therefore (1-c^2) \ge (\frac 75-c)^2$$ $$\therefore \frac 34 \le c \le \frac 45$$

How can I use lagrange multiplier method on this problem effectively? Thanks.

Answer 1

You’ve already got everything you need. You can either proceed with two multipliers: $$L(a,b,c;\lambda,\mu) = c - \lambda(a^2+b^2+c^2-1) - \mu\left(b+c-\frac75\right)$$ or incorporate the linear constraint right away: $$L(a,c;\lambda) = c-\lambda\left(a^2+\frac1{25}(7-5c)^2+c^2-1\right).$$ I’d do the latter since it immediately eliminates two variables from the calculations and gets you to a set of equations that you’d have arrived at anyway. After differentiating, besides the constraint equation we have $$-2\lambda a = 0 \\ \left(\frac{14}5-4c\right)\lambda+1 = 0.$$ From this we quickly find that $\lambda\ne0$, so $a=0$, which leaves a quadratic equation in $c$.