Question

The sum of two numbers is fixed. Then its multiplication is maximum, when

• A. Each number is half of the sum.
• B. Each number is $\frac{1}{3}$ and $\frac{2}{3}$ respectively of the sum.
• C. Each number is $\frac{1}{4}$ and $\frac{3}{4}$ respectively of the sum.
• D. None of the above

Answer 1

The correct option is A

Each number is half of the sum.

Explanation for the correct option:

Let us assume that $x$ and $y$ are the two numbers and let $s$ be their sum. Thus,

$$\begin{gathered} x+y=s \\ \Rightarrow y=s-x...\left(\mathrm{i}\right) \end{gathered}$$

Let the multiplication of the $x$ and $y$ be denoted as $f\left(x\right)$

Thus, $f\left(x\right)=xy$

We know that when the derivative of a function is $0$, it is a local maxima or minima i.e., it attains the highest or lowest value around a given range, respectively.

Thus, by differentiating $f\left(x\right)$ and setting it equal to $0$, we get

$\begin{array}{ccc}& f\text{'}\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}xy& \\ \Rightarrow & 0=\frac{\mathrm{d}}{\mathrm{d}x}x\left(s-x\right)& \left[\because y=s-x\right]\\ \Rightarrow & 0=\frac{\mathrm{d}}{\mathrm{d}x}\left(xs-x^2\right)& \\ \Rightarrow & 0=s-2x& \\ \Rightarrow & x=\frac{s}{2}& \end{array}$

Substituting the above value in equation $\left(\mathrm{i}\right)$,

$$\begin{gathered} y=s-\frac{s}{2} \\ \Rightarrow y=\frac{s}{2} \end{gathered}$$

Thus, $x=y=\frac{s}{2}$

Hence, $\mathrm{Option}\left(\mathrm{A}\right)$is correct.

Note:

Why is it a maxima and not a minima? Because, if you see the third equation in the differentiating part, the equation on the right side being differentiated is $xs-x^2$. It is a quadratic equation, a parabola, with a negative sign on the quadratic term. This means that the parabola opens downward and has the highest value at the tip.