Question

How do you find the two positive real numbers whose sum is 40 and whose product is maximum.

Answer 1

Hint : Here in this question, we have to read the question and find the relation by reading the question. By knowing the relation, we can have to write in the form of numerals. While writing in the numeral form we use the variables, constants and arithmetic operations. After that we have to determine the values.

Complete step-by-step answer :
From the given data of a question, we have to find the two numbers which are real and they are positive.
So let we consider the two real and positive numbers to be “x” and “y”.
In the question they have given the sum of these two numbers is 40. Therefore, we have $x + y = 40$----(1)
The product of these two numbers is maximum. therefore, we have $x.y$---(2)
From (1) we write the equation as $y = 40 - x$----(3)
When we substitute this in the equation (2) we get
$x(40 - x)$, let we consider it as $f(x)$. Therefore, the given equation can be written as
$\Rightarrow f(x) = x(40 - x)$
On simplifying we write it as
$\Rightarrow f(x) = - {x^2} + 40x$
Since the product is maximum, we differentiate the above function. On differentiating we get
$\Rightarrow f'(x) = - 2x + 40$
Again we differentiate the function we get
$\Rightarrow f''(x) = - 2$
When we differentiate the function we obtain the negative. so we not consider the double differentiation. We consider only the first differentiation. i.e., $f'(x) = - 2x + 40$ since it is maximum the value of $f'(x) = 0$. Therefore we have
$\Rightarrow 0 = - 2x + 40$
Take -2x to LHS we get
$\Rightarrow 2x = 40$
On dividing the above equation by 2 we get
$\Rightarrow x = 20$
On substituting the x value in equation (3) we get
$y = 40 - 20$
On simplifying we get
$y = 20$
Therefore the two positive real numbers are 20 and 20.
So, the correct answer is “Therefore the two positive real numbers are 20 and 20.”.

Note : Since in question they have given the product is maximum we use the differentiation topic, where it is one of the applications of derivatives. While simplifying we use the simple arithmetic operations like addition, subtraction, multiplication and division.