I was working on a problem:
"Given $x+2y+3z=15$, find the minimal value of $x^2+y^2+z^2$."
I tried to find the answer by balancing out each of the three variables. However, I there appeared to be a smaller value than the answer I found. Does anyone know how I am intended to solve this?
On
We are given that $f(x,y,z)= x^2+y^2+z^2$ with the constraint $x + 2y+3z=15$. Let's define $g(x,y,z)=x + 2y+3z - 15$ and then relate partial derivatives between $f$ and $g$. We will multiply all the partial derivatives of $g$ by our Lagrange multiplier $\lambda$.
$$2x=\lambda$$ $$2y=2\lambda$$ $$2z=3\lambda$$
We now need to solve all three of these equations for $\lambda$. Multiply all of them to get $2xyz$ on the LHS.
$$2xyz = \lambda yz$$ $$2xyz = 2\lambda xz$$ $$2xyz = 3\lambda xy$$
If we divide both sides by $\lambda$ then the LHS for all three equations will be $\dfrac{2xyz}{\lambda}$. Then, all three equations on the RHS are equal and we can write
$$yz=2xz=3xy$$
Therefore, if we review our constraint $x + 2y+3z=15$, we see that we form the following solutions:
$$x=0,~y=0,~z=5\tag{1}$$ $$x=0,~y=\frac{15}{2},~z=0\tag{2}$$ $$x = \frac{15}{14},~ y = \frac{15}{7},~ z = \frac{45}{14}\tag{3}$$ $$x = 15,~ y = 0,~ z=0\tag{4}$$
and if we evaluate these against $f(z,y,z)$
$$f(x,y,z)=25\tag{1*}$$ $$f(x,y,z)=\frac{225}{4}\tag{2*}$$ $$f(x,y,z)=\frac{225}{14}\tag{3*}$$ $$f(x,y,z)=225\tag{4*}$$
we see that $(3)$ returns the minimum value of $f(x,y,z)=\dfrac{225}{14}$.
by cauchy Schwarz we get $$(x+2y+3z)^2\le (x^2+y^2+z^2)(1+4+9)$$ so $$\frac{225}{14}\le x^2+y^2+z^2$$