Question

One number is 4 less than twice a second number. find a pair of such numbers so that their product is as small as possible.

Answer 1

 For this case, the first thing we must do is define a variable.
 We have then:
 x: unknown number.
 First number: 2x - 4
 Second number: x
 The product of the numbers is:
 $y = x (2x-4)$
 Rewriting:
 $y = 2x ^ 2-4x$
 Deriving the equation we have:
 $y '= 4x-4$
 We equal zero and clear x:
 $4x - 4 = 0 x = 4/4 x = 1$
 Then, the first number is:
 $2x - 4 = 2 (1) - 4 = 2 - 4 = - 2$
 Answer:
 First number: -2
 Second number: 1 

Answer 2

Answer:

Numbers are -2 and 1.

Step-by-step explanation:

Let x be the second number,

⇒ First number = 4 less than twice a second number

= 2 × Second number - 4

= 2x - 4

Thus, the product of first and second number is,

$f(x) = x(2x-4)$

$\implies f(x) = 2x^2 - 4x$

Differentiating with respect to x,

$f'(x) = 4x -4$

Again differentiating with respect to x,

$f''(x) = 4$

Now, for maximum or minimum,

$f'(x)=0$

$\implies 4x - 4 = 0\implies 4x = 4\implies x = 1$

Since, for x = 1, f''(x) = Positive,

Therefore, the function f(x) is minimum for x = 1,

⇒ The product is smallest for x = 1,

Hence, the second number = x = 1,

And, first number = 2x - 4 = 2 - 4 = -2