$y=(10-a)(10-b)(10-c)$
$\text{where } (a+b+c)=3\text{ and each of (a,b,c) are positive}$
What is the maximum and minimum value $y$ can have?
I have created this question with these small values for ease of understanding. From my trials in excel, It appears as
Maximum is when $a=b=c=1$. Then, $y=9×9×9=729$
Minimum is when any two of $(a,b,c)$ is zero and other is $3$. Then, $y=10×10×7=700$
So, $700 \lt y \le 729$ is the required range.
But, These are based on my trials and may be wrong also. Please tell whether my answer is right or wrong and also help in deriving this answer mathematically.
Update:wolframalpha.com gives the following results
maximum : 729
https://www.wolframalpha.com/input/?i=maximize+(10-a)(10-b)(10-c)++on+a%2Bb%2Bc%3D3minimum : no global minima found
http://www.wolframalpha.com/input/?i=minimize+(10-a)(10-b)(10-c)++on+a%2Bb%2Bc%3D3
Best Answer
The set $D=\{(a,b,c)\in\mathbb{R}^3:a+b+c=3,a,b,c>0\}$ is not a closed set with a smooth boundary, hence the method of Lagrange multipliers requires to be adjusted. By computing the partial derivatives of $(10-a)(10-b)(10-c)$ and $a+b+c$ we have that $a=b=c=1$ is a stationary point (indeed an absolute maximum) but we still have to study what happens on $\partial D=\{(a,b,c)\in\mathbb{R}^3:a+b+c=3,abc=0\}$. Since $\partial D$ is symmetric with respect to cyclic shifts of the variables, we may assume $a=0$ and study $10(10-b)(10-c)$ under the constraints $b+c=3$ and $b,c\geq 0$. It follows that $\color{red}{700}$ is an infimum over $D$ and $\color{red}{729}$ is a maximum, as conjectured.