Find max/min of $f(x,y,z)$ on closed unit ball $B$ in $\mathbb{R^3}$

Question

$f(x,y,z) = 3x - 2y + z$

Let $B$ be a closed unit ball in $\mathbb{R^3}$, find the max/min of f on $B$.

We first need to observe

$(a)$ the behavior of $f$ in $B^0$

$(b)$ the behavior of $f$ on $\partial B$

Then we can find the $\max_B f$ and $\min_B f$.

I am confused as to how to do this. I think that since this is a smooth function, we know it takes its maximum value on $B$ and the point where the maximum occurs will solve the Lagrange Multiplier equation.

Any help is greatly appreciated!

Answer 1

For $$\vec a = (3, - 2,1),\vec b = \frac{1}{{\left\| {\vec a} \right\|}}\vec a$$ is given: $$f(\vec x) = \vec a \cdot \vec x$$ and constraint condition $$g(\vec x) = {\left\| {\vec x} \right\|^2} - {R^2}$$ We build the Lagrange function: $$L(\vec x,\lambda ) = f(\vec x) + \lambda \cdot g(\vec x)$$ and calculate: $${\nabla L = \left( {\begin{array}{*{20}{c}} {2\lambda \cdot \vec x + \vec a} \\ {{{\left\| {\vec x} \right\|}^2} - {R^2}} \end{array}} \right) = \vec 0}$$ That means $$\begin{array}{*{20}{c}} {2\lambda \cdot \vec x + \vec a = \vec 0} \\ {{{\left\| {\vec x} \right\|}^2} - {R^2} = 0} \end{array}$$ Solution is not hard to find: $$\lambda = - \frac{1}{{2 \cdot R}}\left\| {\vec a} \right\|,{{\vec x}_{\max }} = R\vec b,{{\vec x}_{\min }} = - R\vec b$$ And we have: $$f({{\vec x}_{\max }}) = R\left\| {\vec a} \right\|,f({{\vec x}_{\min }}) = - R\left\| {\vec a} \right\|$$ With $$R \geqslant 0$$ one can study different cases.