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$y=(10-a)(10-b)(10-c)$

$\text{where } (a+b+c)=3\text{ and each of (a,b,c) are positive}$

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%3D3

minimum : no global minima found http://www.wolframalpha.com/input/?i=minimize+(10-a)(10-b)(10-c)++on+a%2Bb%2Bc%3D3

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    $\begingroup$ For max $$\dfrac{10-a+10-b+10-c}3\ge\sqrt[3]{(10-a)(10-b)(10-c)}$$ $\endgroup$ – lab bhattacharjee Aug 30 '16 at 11:38
  • $\begingroup$ @labbhattacharjee, is it based in AM $\ge$ GM $\endgroup$ – Kiran Aug 30 '16 at 11:40
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    $\begingroup$ Did you already learn Lagrange multipliers ? $\endgroup$ – Claude Leibovici Aug 30 '16 at 11:41
  • $\begingroup$ @Claude Leibovici, sorry, i don't know that. $\endgroup$ – Kiran Aug 30 '16 at 11:41
  • $\begingroup$ Don' worry ! This will be soon and I am sure that you will enjoy them. Continue making problems for your fun and our entertainment ! Cheers. $\endgroup$ – Claude Leibovici Aug 30 '16 at 16:24
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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.

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Since you did read the Wikipedia page, you will not have any problem to follow what I am writing here.

We want to find maximum and minimum of the function $$y=(10-a)(10-b)(10-c)$$ subject to the constraint $(a+b+c)=3$. So, let us define $$F=(10-a)(10-b)(10-c)+\lambda(a+b+c-3)$$ which is the function to minimize/maximize.

Compute the derivatives $$F'_a=\lambda -(10-b) (10-c)\tag 1$$ $$F'_b=\lambda -(10-a) (10-c)\tag 2$$ $$F'_c=\lambda -(10-a) (10-b)\tag 3$$ $$F'_\lambda=a+b+c-3\tag 4$$ AT the extremum, these derivatives will all be equal to zero.

Looking at $(1)$ and $(2)$, it is clear that $a=b$; looking at $(1)$ and $(3)$, it is clear that $a=c$. So, $a=b=c$ and the constraint implies $a=b=c=1$; all of that reduces the problem to $$y=(10-a)^3$$ to be maximized/minimized for $0<a \leq 1$. From there, your result.

Edit

The problem would have been slightly more difficult considering $$y=(A-a)(B-b)(C-c)$$ subject to the constraint $(a+b+c)=D$. Doing the same as above $$F'_a=\lambda -(B-b) (C-c)\tag 5$$ $$F'_b=\lambda -(A-a) (C-c)\tag 6$$ $$F'_c=\lambda -(A-a) (B-b)\tag 7$$ $$F'_\lambda=a+b+c-D\tag 8$$ Eliminating $a,b,c$ from $(5),(6),(7)$ we should obtain $$a=A\pm \sqrt \lambda \qquad b=B\pm \sqrt \lambda \qquad c=C\pm \sqrt \lambda$$ and using $(8)$, this would lead to $$A+B+C\pm 3\sqrt \lambda=0\implies \lambda=\pm \frac {A+B+C}3$$ then $a,b,c$.

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