How do you find the two numbers whose difference is 40 and whose product is a minimum ?

Hint:To solve the given question, first we need to let the two numbers according to the question given. And it is given that the product of two numbers is minimum for that we will find the product of two numbers and then differentiate it with respect to ‘x’. Then to find the minimum value we will equate the differentiate value equal to 0 and will get the value of ‘x’ and the other number.

Complete step by step answer:
We have given that the difference of two numbers is 40. Therefore,let the first number be \[x\].And the other number is \[x+40\] (as it is given in the question that the difference is 40.) According to the question;
The products of two numbers are,
\[ x\left( x+40 \right)\]
By using the distributive property of multiplication, i.e. \[a\left( b+c \right)=ab+ac\]
Thus,
\[ x\left( x+40 \right)={{x}^{2}}+40x\]

According to the question, it is given that the product of two numbers is a minimum.The above value will have minimum value when the value of ‘x’ is greater than 0. This implies that the coefficient of \[{{x}^{2}}\]should be 1. Thus, equating the product of two numbers equals to 0.
\[{{x}^{2}}+40x=0\]
Derivative of the above expression, we obtain
\[\dfrac{d}{dx}\left( {{x}^{2}}+40x \right)=2x+40\]
Thus, the minimum occurs at,
\[2x+40=0\]

Now, finding the value of ‘x’,
Subtracting 40 from both the sides of the equation, we get
\[2x=-40\]
Dividing both the sides of the equation by 2, we get
\[x=-20\]
Therefore, the two number are,
\[x=-20\], and\[x+40=-20+40=20\].
Thus,
\[\Rightarrow x\left( x+40 \right)=\left( -20 \right)\times 20\\
\therefore x\left( x+40 \right)=-400\]

Hence, the two numbers are -20 and 20 whose difference is 40 and product is a minimum.

Note:While solving these types of questions, students need to let the numbers by considering the conditions given in the question. As it may have seen many times they let wrong numbers and made mistakes and gave the answer as undetermined. Students need to be very careful while doing the calculation part to avoid making errors.