Question

Find two numbers whose difference is 80 and whose product is a minimum.

Answer 1

Answer:

The numbers are (40,-40).

Step-by-step explanation:

Given : Two numbers whose difference is 80 and whose product is a minimum.

To find : The numbers ?

Solution :

Let the largest number be x,

and smallest number be y.

According to question,

Two numbers whose difference is 80.

i.e. $x-y=80$ ......(1)

Product of two number is minimum.

i.e. $xy=P$ .....(2)

where P is minimum.

Now substitute x from (1) in (2),

$x(x-80)=P$

$x^2-80x=P$

Apply completing the square by adding 40 square both side,

$x^2-80x+(40)^2=P+(40)^2$

$(x-40)^2=P+(40)^2$

The vertex of the equation occur at x=40.

So, the value of y is $40-y=80$

$y=-40$

Therefore, the numbers are (40,-40).

Answer 2

See the suggested solution in the attachment.
Answer: -40 and 40.