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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.

Respuesta :

carlosego
 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:
 [tex] y = x (2x-4) [/tex]
 Rewriting:
 [tex] y = 2x ^ 2-4x [/tex]
 Deriving the equation we have:
 [tex] y '= 4x-4 [/tex]
 We equal zero and clear x:
 [tex] 4x - 4 = 0 x = 4/4 x = 1[/tex]
 Then, the first number is:
 [tex] 2x - 4 = 2 (1) - 4 = 2 - 4 = - 2 [/tex]
 Answer:
 First number: -2
 Second number: 1 

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,

[tex]f(x) = x(2x-4)[/tex]

[tex]\implies f(x) = 2x^2 - 4x[/tex]

Differentiating with respect to x,

[tex]f'(x) = 4x -4[/tex]

Again differentiating with respect to x,

[tex]f''(x) = 4[/tex]

Now, for maximum or minimum,

[tex]f'(x)=0[/tex]

[tex]\implies 4x - 4 = 0\implies 4x = 4\implies x = 1[/tex]

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

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