Greatest product (Using Calculus)

Question

We know that whatever number $n$ may be, we must divide it into two equal parts if the product of the parts is to be a maxi- mum; and the value of that maximum product will always be $= 0.25n^2$.

Let the number to be cut into two parts be called n. Then if $x$ is one part, the other will be $n - x$, and the product will be $x(n - x)$ or $nx - x^2$ • So we write $y = nx - x^2$ • Now differentiate and equate to zero; We get, $n-2x=0$ or $n/2=x$. I know it worked for two parts.

Now, my question is, how it would work for proving that the product is max. if these three numbers($m,n,p$) are equal, when sum of these three is constant. we are given three numbers, such that $m+n+p=k$, where $k$ is some constant. $(m,n,p,)$ are positive real numbers.

Thanks in advance.

Answer 1

The easiest way to go about your problem is the AM-GM inequality, which tells you that, for any set of positive real numbers $x_1,\dots, x_n$, the inequality

$$\frac{x_1+\cdots +x_n}{n}\geq \sqrt[n]{x_1\cdots x_n}$$

is true.

In your case, we know that $x_1+\cdots + x_n=k$, which means that $\sqrt[n]{x_1\cdots x_n}\leq \frac{k}{n}$, and therefore $x_1\cdots x_n \leq \frac{k^n}{n^n}, is always true.

Selecting $x_i=\frac{k}{n}$ also means that

$$x_1\cdots x_n = \frac{k^n}{n^n}$$

which means that $\frac{k^n}{n^n}$ is the maximum of the expression $x_1\cdots x_n$ under the conditions $x_1,\dots x_n\geq 0$ and $x_1+\cdots + x_n = k$.

Answer 2

Given two numbers $y\geq x>0$ with sum $n$ you can write them as $$x={n\over2}-h,\quad y={n\over2}+h$$ with some $h\geq0$. It follows that $$xy={n^2\over4}-h^2<{n^2\over4}\qquad(h>0)\ .$$ This shows that $x=y={n\over2}$ gives the maximal product.

If we are given $r$ numbers $x_i>0$ $(1\leq i\leq r)$ with sum $n$, and two of them are unequal, say $x_1<x_2$, then we can enlarge the product $\prod_{i=1}^n x_i$ by replacing $x_1$ and $x_2$ with $x_1'=x_2'={x_1+x_2\over2}$. Doing this we have not alered the sum $n$ of all ixes. If you accept without proof that a maximal product existsthis maximal product must have all ixes equal, namely $x_i={n\over r}$ $(1\leq i\leq r)$.

Answer 3

Let $x_1, x_2, \ldots, x_n \ge 0$ are nonnegative reals, subject to the constraint $x_1 + x_2 + \cdots + x_n = k$. We reason by induction that the maximum value of the product $x_1 x_2 \ldots x_n$ is attained when $x_1 = x_2 = \ldots = x_n = k/n$.

First, the case where $n = 2$ was already addressed. So in the case $n = 3$, suppose $x_3$ is fixed, hence the maximum occurs when $x_1 = x_2 = (k - x_3)/2$ and the resulting product is $x_1 x_2 x_3 = (k - x_3)^2 x_3/2$. Now if $x_3$ is allowed to take on different values in $[0,k]$, the fact that the product is maximized when $x_1 = x_2$ remains true, so we can argue that as a function of $x_3$, the product is maximized for some critical point of this function, namely $$0 = \frac{d}{dx_3}\left[\frac{(k - x_3)^2 x_3}{2}\right] = (k - x_3)x_3 + \frac{(k - x_3)^2}{2}.$$ Solving the quadratic results in $x_3 = k$ (which obviously corresponds to a minimum) and $x_3 = k/3$. A convexity argument indicates this yields the desired maximum, and $x_1 = x_2 = x_3 = k/3$.

The above is naturally extensible. For suppose that we have established that the product is maximized on $n$ variables when all are equal to $k/n$. Then for $n+1$ variables, consider $x_{n+1}$ fixed, so the product is $$x_1 x_2 \ldots x_n x_{n+1} = \frac{(k-x_{n+1})^n x_{n+1}}{n}.$$ Searching for the critical point with respect to $x_{n+1} \in [0,k]$ yields $x_{n+1} = k/(n+1)$, hence $x_1 = x_2 = \ldots = x_{n+1} = k/(n+1)$ and this completes the induction step.

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Now, if $x_1, \ldots, x_n$ are nonnegative integers, the above reasoning does not work because we cannot compute critical points to obtain extrema. Instead, we reason more simply. Suppose $x_1 + \cdots + x_n = k$. Then adding $1$ and subtracting $1$ from $x_i$ and $x_j$ respectively, for $i < j$ (without loss of generality), yields the same sum, but the product becomes $x_1 x_2 \ldots x_n$ to $x_1 \ldots (x_i + 1) \ldots (x_j - 1) \ldots x_n$. If we take the ratio, we find this is $$\frac{(x_i + 1)(x_j - 1)}{x_i x_j} = \frac{x_i x_j - x_i + x_j - 1}{x_i x_j} = 1 - \frac{1}{x_j} + \frac{1}{x_i} - \frac{1}{x_i x_j}.$$ So if $x_i = x_j - 1$, this ratio equals $1$, and the value of the product remains unchanged. From this, we can see that if $x_i \ge x_j$, then incrementing $x_i$ by $1$ and decrementing $x_j$ by $1$ will decrease the product, and if $x_i < x_j - 1$, it will increase the product. This is irrespective of the values of the other factors in the product.

Hence, if there exists $i \ne j$ such that $x_i < x_j - 1$, repeatedly incrementing $x_i$ and decrementing $x_j$ so that $|x_i - x_j| \le 1$ will increase the value of the product. Again without loss of generality, order the terms in nondecreasing sequence, so that $x_i \le x_j$ for all $i < j$. Then we can see that we must have $|x_n - x_1| \le 1$, which implies that there is a $m$ such that $x_1 = \ldots = x_m$, and $x_{m+1} = \ldots = x_n = x_m + 1$, or $$mx_1 + (n-m)(x_1+1) = (n-m) + nx_1 = k.$$ Since all of these variables are nonnegative integers, we require $n$ to divide $k+m$. Since $1 \le m \le n$, such a choice of $m$ is unique as there is only one representative of the equivalence class of integers modulo $n$ that is the additive inverse of $k$, namely $m \equiv -k \pmod n$. As this choice is unique, the corresponding product must be the maximum attainable over the nonnegative integers.

An example of this maximization for $n = 7$, $k = 45$: we select $m \in \{1, \ldots, 7\}$ such that $45+m$ is divisible by $7$. This of course is $m = 4$, hence $x_1 = x_2 = x_3 = x_4$ and $x_5 = x_6 = x_7 = x_1 + 1$. We must have $7x_1 + 3 = 45$, or $x_1 = 6$, and the maximum product is $6^4 7^3$.

Answer 4

You can prove AM-GM with calculus (and induction).

Let's prove that, given $x_1,\dots,x_k>0$, we have $$ \sqrt[k]{x_1x_2\dotsm x_k\mathstrut}\le \frac{x_1+x_2+\dots+x_k}{k} $$ For simplicity, denote by $\gamma(x_1,\dots,x_k)$ the left-hand side and by $\alpha(x_1,x_2,\dots,x_k)$ the right-hand side. Thus we want to prove $$ (\text{AM-GM})_k\qquad \gamma(x_1,\dots,x_k)\le\alpha(x_1,\dots,x_k) $$

The base case of induction, $k=1$, is obvious. Suppose we know the statement for $k$ and we're given $k+1$ positive numbers $x_1,\dots,x_k,x$; we want to prove that $$ \gamma(x_1,\dots,x_k,x)\le\alpha(x_1,x_2,\dots,x_k,x) $$ which is equivalent to $$ (k+1)^{k+1}\gamma(x_1,\dots,x_k,x)^{k+1}\le\alpha(x_1,x_2,\dots,x_k,x)^{k+1} $$ Let's write $P=(x_1\dotsm x_k)^{1/k}$ and $S=x_1+\dots+x_k$ so the statement to be proved becomes $$ (k+1)^{k+1}P^kx\le(S+x)^{k+1} $$ under the induction hypothesis that $kP\le S$. Consider the function $$ f(x)=(S+x)^{k+1}-(k+1)^{k+1}P^kx $$ Then $f'(x)=(k+1)(S+x)^k-(k+1)^{k+1}P^k$, which vanishes for $S+x=(k+1)P$, that is, $x=(k+1)P-S$. Note that $$ \lim_{x\to0}f(x)=S^{k+1}>0,\qquad \lim_{x\to\infty} f(x)=\infty $$ If $(k+1)P-S\le0$, the function is increasing and so the inequality is proved. Suppose $(k+1)P-S>0$. Then $f$ has an absolute minimum at $(k+1)P-S$ and \begin{align} f((k+1)P-S) &=(k+1)^{k+1}P^{k+1}-(k+1)^{k+1}P^k(kP+P-S)\\[4px] &=(k+1)^{k+1}P^k\bigl(P-kP-P+S\bigr)\\[4px] &=(k+1)^{k+1}P^k(S-kP)\ge0 \end{align} Thus the inequality is proved in every case.

Notice also that the minimum is zero if and only if $S=kP$. Thus, again by induction, we prove that equality in AM-GM$_k$ is attained only when $x_1=x_2=\dots=x_k$.

With the inequality at hand, assume $x_1+\dots+x_k=n$. Then $$ x_1x_2\dotsm x_k\le(n/k)^k $$ On the other hand, for $x_1=x_2=\dots=x_k$, the two terms are equal and equality only holds in this case.